LABORATORY MANUAL OF GENERAL 

CHEMISTRY 

WITH 




BY 

ARTHUR B. LAMB 

ASSISTANT PROFESSOR OF CHEMISTRY AND DIRECTOR OF THE 
CHEMICAL LABORATORY OF HARVARD UNIVERSITY 




CAMBRIDGE 

HARVARD UNIVERSITY PRESS 
1916 



COPYRIGHT, 1916 
HARVARD UNIVERSITY PRESS 



PREFACE 

THIS laboratory manual has been written to meet the requirements of students 
of chemistry who already possess an elementary knowledge of the subject, such, 
for instance, as is acquired at our better high schools. 

How best to continue the chemical education of such students is one of the 
most difficult problems which confront teachers of chemistry in our colleges and 
universities. A time honored practice has been to ignore secondary school 
preparation entirely and give identical instruction to these men and to real be- 
ginners indiscriminately. This was doubtless justifiable some years ago when 
chemical instruction was new in our secondary schools and was, naturally, poor; 
but today it can only be defended on the ground of necessity. The larger insti- 
tutions recognize this where the number of elementary students is adequate, and 
either have arranged separate laboratory sections for the differently prepared 
students, or, better still, give them wholly separate instruction. 

But even when segregation of this kind has been secured, the problem is by 
no means solved. Those students who have studied chemistry in the secondary 
school have already done a large share of the simple, important, and impressive 
experiments. The first freshness of their interest in, and wonder at, chemical 
phenomena has been lost. On the other hand, to trust that the average college 
student retains any clear conceptions regarding the abstract matter of his second- 
ary school chemistry, which is so important as a basis for further study, is to 
court disappointment. Besides, as every experienced college teacher knows, the 
very familiarity of such students with parts of the subject frequently leads to 
over-confidence about the whole of it with disastrous results. Flagging 
interest then, hazy ideas about the principles of chemistry, and over-confidence 
are the special difficulties of the problem. 

The requirements then to be met by a laboratory manual of this kind are by 
no means easy. The most essential are, first, that those important facts and 
principles which the student has already studied shall be reviewed in a way suffi- 
ciently novel not to bore him, nor to encourage him to over-confidence; second, 
that the student's chemical horizon shall be widened by the study of new and 
unfamiliar substances; and third, that further important generalizations upon 
which the superstructure of the science is based, shall be disclosed and made 
clear. 

To meet these requirements I have resorted to several expedients. For in- 
stance, to review the weight relationships of chemical reactions, I have devised 
a series of simple, quantitative experiments quite different from the ones usually 
performed in a strictly elementary course. A number of new experiments of a 



358843 



PREFACE 

semi-quantitative nature have also been devised, illustrating such matters as the 
vapor pressure of solutions, reaction velocity, catalysis, the strength of oxidizing 
agents, etc. 

Again, the descriptive experiments have been correlated, when possible, with 
some general principle; that is, they have been made deductive as well as induc- 
tive. Finally, a relatively large number of inorganic preparations have been 
introduced in order to increase the student's familiarity with chemical sub- 
stances and chemical operations, and to take advantage of that feeling of satis- 
faction which most students experience in actually making a substance. 

For convenience, these preparations have been grouped together as Part II 
of the manual, but in actual practice they are interpolated at appropriate points 
among the other experiments. 

Everywhere, I have attempted, for educational reasons, to emphasize the 
logical side of the subject. I have also prefaced nearly every experiment with a 
paragraph or so, pointing out the object and the bearing of the experiment. 
This has been done both in the hope of discouraging a merely mechanical exe- 
cution of the experiment, and in the belief that the more the student under- 
stands what he is to look for, the more he will observe ; in other words, that it 
is the mind and not the eye which sees. 

Finally, I have avoided the insertion of tests and precipitation experimenst 
where these have no significance beyond their use in systematic analysis. These 
matters can be more easily and economically considered in a study of qualitative 
analysis. 

Only an exceptional student could complete all of the experiments and 
preparations included in this manual during the thirty periods of three hours 
allotted to this subject here, and actually the student is required to complete 
only about four-fifths of them; the others, which include the less important or 
more difficult ones, are given out as " optional " experiments or preparations to 
the ambitious or. specially interested student. 

In working over the preparations, I found the small manual by Blanchard, 
entitled " Synthetic Inorganic Chemistry," and the larger one by Biltz, entitled 
" Laboratory Methods of Inorganic Chemistry," especially helpful. I have made 
use of many of their suggestions. I wish also to acknowledge gratefully the 
valuable assistance of Mr. Willis A. Boughton and Professor Elmer P. Kohler, 
of this Laboratory, in the preparation of the manuscript and in the reading of 
the proofs. 

ARTHUR B. LAMB. 

CHEMICAL LABORATORY OF HARVARD UNIVERSITY, 
CAMBRIDGE, MASS., September, 1916. 



TABLE OF CONTENTS 



PREFACE iii 

GENERAL INSTRUCTIONS FOR LABORATORY WORK . 3 

PART I. EXPERIMENTS IN GENERAL CHEMISTRY 

EXPERIMENT 

I. To DETERMINE THE PROPORTION BY WEIGHT IN WHICH OXYGEN AND COPPER 

COMBINE TO FORM CUPROUS OXIDE 7 

II. To DETERMINE THE PROPORTION BY WEIGHT IN WHICH SULPHUR AND COPPER 

COMBINE TO FORM CUPROUS SULPHIDE . . 9 

III. To DETERMINE THE PROPORTION BY WEIGHT IN WHICH OXYGEN AND SULPHUR 
COMBINE TO FORM SULPHUR DIOXIDE 11 

IV. OXIDATION AND REDUCTION " 13 

V. HYDROGEN PEROXIDE. PEROXIDES 15 

VI. AUTOXIDATION 17 

VII. SOLUTIONS. GASES IN LIQUIDS. To DETERMINE THE SOLUBILITY OF AIR IN 

WATER 19 

VIII. SOLUTIONS. SOLIDS IN LIQUIDS. To DETERMINE THE SOLUBILITY OF A SOLID 

IN A LIQUID ! 21 

IX. SOLUTIONS. SOLIDS IN LIQUIDS (CONTINUED). To OBSERVE THE EFFECT OF 

TEMPERATURE ON SOLUBILITY 21 

X. SUPERSATURATED SOLUTIONS 23 

XI. SOLUTIONS. To OBSERVE THE EFFECT OF DISSOLVED SOLIDS ON THE VAPOR 

PRESSURE 23 

XII. SOLUTIONS. To OBSERVE THE EFFECT OF DISSOLVED SOLIDS ON THE BOILING 

POINT 27 

XIII. THE VELOCITY OF CHEMICAL REACTION 29 

XIV. CHEMICAL EQUILIBRIUM 31 

XV. THE ELECTRICAL CONDUCTIVITY OF SOLUTIONS 33 

XVI. IONIC EQUILIBRIA 37 

XVII. THE TITRATION OF ACIDS AND BASES , 39 

XVIII. HYDROGEN SULPHIDE . 41 

XIX. HYDROLYSIS 43 

XX. SULPHUROUS ACID 45 

XXI. SULPHURIC ACID 45 

XXII. CATALYSIS 47 

XXIII. PROPERTIES OF FREE BROMINE AND IODINE 49 

XXIV. PREPARATIONS AND PROPERTIES OF HYDROGEN FLUORIDE, BROMIDE, AND IODIDE 51 
XXV. PRECIPITATION REACTIONS OF HALOGEN IONS 53 

XXVI. THE RELATIVE AFFINITIES OF THE HALOGENS FOR NEGATIVE ELECTRICITY . . 53 

XXVII. THE ABSORPTION OF IODIDES BY THE STOMACH 55 

XXVIII. THE OXYGEN ACIDS OF THE HALOGENS 55 

XXIX. THE RELATIVE OXIDIZING POWERS OF THE OXYGEN ACIDS OF THE HALOGENS . 57 

XXX. AMMONIA AND AMMONIUM HYDROXIDE .' 59 

XXXI. NITRIC ACID AS AN OXIDIZING AGENT 61 

XXXII. PHOSPHINE 63 

XXXIII. THE PHOSPHORIC ACIDS 63 

XXXIV. PHOSPHOROUS ACID . 65 



TABLE OF CONTENTS 

EXPERIMENT 

XXXV. CARBON MONOXIDE 67 

XXXVI. SATURATED AND UNSATURATED H^ROCARBONS 67 

XXXVII. ALCOHOLS 69 

XXXVIII. ORGANIC ACIDS .'................ 71 

XXXIX. BORON AND SILICON 73 

XL. ALKALINE EARTH ELEMENTS 75 

XLI. MAGNESIUM 77 

XLII. ZINC, CADMIUM, AND MERCURY 79 

XLIII. OVERVOLTAGE . " 81 

XLIV. ALUMINUM 83 

XLV. ARSENIC, ANTIMONY, AND BISMUTH 85 

XLVI. TIN AND LEAD 87 

XLVII. COLLOIDAL SOLUTIONS 89 

XLVIII. MANGANESE 91 

XLIX. CHROMIUM 93 

L. SILVER 95 

PART II. EXERCISES IN THE PREPARATION OF INORGANIC SUBSTANCES 

DIRECTIONS FOR LABORATORY WORK 97 

EXERCISE 

I. POTASSIUM NITRATE FROM SODIUM NITRATE AND POTASSIUM CHLORIDE ... 99 

II. PURE SODIUM CHLORIDE FROM ROCK SALT 103 

III. STRONTIUM CHLORIDE FROM CELESTITE (STRONTIUM SULPHATE) 107 

IV. AMMONIUM IRON ALUM FROM SIDERITE Ill 

V. AMMONIO-CUPRIC SULPHATE 115 

VI. AMMONIUM BROMIDE 117 

VII. ARSENIC ACID 121 

VIII. CUPROUS CHLORIDE FROM COPPER 125 

IX. MERCURIC CHLORIDE, HgCl 2 (CORROSIVE SUBLIMATE) 129 

X. HYDRAZINE SULPHATE 131 

XI. MERCURIC THIOCYANATE 135 

XII. CHLORO PENTAMMINE COBALTIC CHLORIDE 137 

XIII. POTASSIUM IODATE 141 

XIV. BARIUM DITHIONATE 143 

XV. CHROMOUS ACETATE 145 

XVI. POTASSIUM THIOCARBONATE 149 

XVII. SELENIUM AND TELLURIUM FROM TELLURIUM SLAG 151 

APPENDIX 155 

APPARATUS REQUIRED FOR EXPERIMENTS AND EXERCISES IN THIS MANUAL .... 156 

ATOMIC WEIGHTS AND LOGARITHMS OF ATOMIC WEIGHTS 158 

PERIODIC CLASSIFICATION OF THE ELEMENTS 159 

LOGARITHMS FOURPLACE TABLE 160 

LOGARITHMS OF CHEMICAL FACTORS 162 

INDEX . 163 



LABORATORY MANUAL OF GENERAL 

CHEMISTRY 



LABORATORY MANUAL OF GENERAL 

CHEMISTRY 

GENERAL INSTRUCTIONS FOR LABORATORY WORK 

The Laboratory Manual. --The directions for each experiment usually 
consist of the following parts : - 

1. Assigned and Suggested Reading; 

2. Discussion of the general principles involved in the experiment; 

3. Procedure, or Directions, for the actual manipulations; 

4. Tests; 

5. Questions. 

The Reading. -- The Assigned Reading is to be done outside of the labora- 
tory; the assignments will usually be found in the text book by Alexander Smith, 
entitled " Introduction to Inorganic Chemistry." Occasionally reading will be 
assigned in other books or periodicals, and this will always be the case with the 
Suggested Reading (optional) . Copies of such books or periodicals will be avail- 
able for reference in the Division Library in Boylston Hall, and usually in the 
Reading Room of the Widener Library. The following abbreviations will be used 

for some of the books most frequently referred to : 

t 

Smith Introduction to Inorganic Chemistry: by Alex- 

ander Smith; New York, 1913. 

Thorp Outlines of Industrial Chemistry: by F. H. 

Thorp; 3d Edition. 1916. 

Roscoe and Schorlemmer Treatise on Chemistry: by Roscoe and Schor- 

lemmer; London, 1905. 

References to Smith in the body of the text are indicated in parenthesis by 
"R" followed by page numbers. 

The Laboratory Work. -- The Discussion accompanying each experiment is 
to be read through carefully before the actual laboratory work is commenced. 
It will save time if this is done outside of the laboratory. The Procedure or the 
Directions should be followed rigorously, as no unnecessary manipulation has 
been included. 

When the word (hood} appears, the operation is not to be conducted at 
the desk in the open laboratory. The apparatus must at once be transferred to 
the hood provided for operations involving ill-smelling gases or vapors. 

3 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

When exact quantities are not indicated, very small amounts of solu- 
tions (1 c.cm: or less} should be taken. This advice is given, partly to secure 
saving of material, but chiefly to avoid the waste of time which working with 
large quantities alwaj^s entails. 

In the first three experiments and in Experiments VIII and XI which are 
quantitative in nature, use the finer balances; in all other cases use the rough 
scales in the laboratory. Before using the finer balances read carefully the direc- 
tions posted on each balance case. If these balances appear to be out of order in 
any way, do not attempt to repair them yourself, but bring the matter to the 
attention of the instructor. 

Chemical Supplies. The substances required for the experiments will be 
found on the shelves at the individual work-desks, or on the shelves at the 
end of each aisle. 

The substances on the shelves at each work-desk consists of aqueous solu- 
tions of sodium hydroxide, ammonium hydroxide, and sulphuric, hydrochloric, 
and nitric acids both concentrated and dilute. These solutions, except the 
ammonium hydroxide, are all of so-called " commercial " purity. The dilute 
solutions are all 6 molar and ought really to be called diluted rather than dilute. 
Each bottle has a labeled place on the shelves; form the habit at the very start 
of returning it to its place after use. This habit will save you time and effort. 

The substances on the shelves at the ends of the aisles are divided into two 
sets, each arranged alphabetically according to their scientific names, The first 
set consists of solids, the second of liquids. Each of these bottles, too, has a 
labeled place, and scrupulous care must be taken to return it to its place. Read 
the labels attentively, as frequently there are several kinds of the saniQ sub- 
stance; for instance,' "pure" and " commercial"; " concentrated" and " dilute," 
etc. The dilute solutions, for the most part, are of molecular normal concen- 
tration. Do not use the pure variety of a substance unless it is specifically called 
for in the directions, as such use would be a needless waste. 

To obtain these substances do not carry the bottles from the end-desks to 
your work-desk; instead, bring to the end-desk a small, clean test-tube for 
liquids, and a clean watch-glass for solids. If too much of any substance is 
taken, do not return it to the bottle, if it could by any possibility have become 
contaminated. 

All materials are supplied through the store-room service. Do not therefore 
take empty bottles to the instructor for refilling, but take them to the 
store-room. 

Notes. Careful and permanent notes are to be made on each experi- 
ment performed. These are to be made in the laboratory, in pencil, on the 
blank page following each experiment and, if necessary, on additional sheets 
provided for that purpose, either during an experiment or immediately after 
its completion. A carbon copy must be secured at the same time. The original 
copy of these notes is to be left in the laboratory in the box provided for that 



purpose, at the completion of the laboratory period. The carbon copy should be 
retained by the student. 

In these notes state first what you did, if anything beyond the directions, 
but do not duplicate the printed directions themselves. Second, state what you 
observed; here your statements cannot be too careful or complete. Third, state 
what conclusions you drew. If the apparatus is at all complicated, a sketch of it 
will enable you to recall the circumstances of the experiment more readily if later 
reference to it is necessary. Whenever a chemical change has been observed the 
equation should, if possible, be given in the notes, but an equation alone is never 
a sufficient record. The " (?) " indicates something to be observed or answered, 
and recorded. The numerous questions asked in the directions themselves are 
intended to be answered in the laboratory, by careful reasoning from the observa- 
tions made. Sometimes the student will find it necessary to carry out further 
tests or experiments of his own before a satisfactory answer is obtained. Do not 
consult an instructor until you have exhausted every means for answering the 
questions independently. 

Reports. The notes made in the laboratory are necessarily brief and dis- 
connected. They are to be supplemented by full and connected reports based 
on these notes and written outside of the laboratory. These reports are to be 
written on the heavy, ruled paper provided for that purpose, and the reports 
for each week are to be bound separately in a folder which also is provided. In 
addition to a connected account of your observations and conclusions, the reports 
are to contain equations for all of the important reactions involved; these are 
indicated in the directions at the point where they occur by "(E)" with an 
appropriate subscript. The reports should also contain answers to the questions 
asked in the directions themselves and to those titled " Outside Questions." 
Answers to these latter are not to be attempted in the laboratory. For those 
experiments devoted to the preparation of substances, special Report-cards are 
to be used, a separate card for each preparation; and a supply of them will 
be furnished at the beginning of the course. The Report-card is to be handed 
to the instructor together with the substance prepared. 

Economy of Time. In laboratory work of this kind there are many un- 
avoidable waits for evaporations, crystallizations, filtrations, etc. Do not waste 
this time in idleness. Indeed, your success in completing the course easily in 
the regular hours will depend on your ability to make each moment count. At 
such times, therefore, either start another experiment, or, remembering that you 
will save time in the end by thoroughly understanding what you are about, study 
the discussions, the questions, or the writing of the equations. 

Accidents. An emergency kit is provided in each laboratory for use in case 
of accident. Find out during the first period where the one in your laboratory is 
located. 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Burns whether caused by hot objects, acids, or corrosive liquids, are quickly 
washed under the tap, if necessary, and then covered with carron oil (an emul- 
sion of lime water and linseed oil). Afterward, all burns, even the slightest, 
should be dressed with the boric acid solution. . 

If anything is spattered in the eye hold it wide open under running water 
from the tap at once. The drinking fountain in the hood at the left of the en- 
trance is most convenient for this purpose. Afterwards rinse the eye with the 
boric acid solution. 

Cuts should be washed in running water and then in the hydrogen peroxide 
solution furnished in the emergency kit, and afterward dressed with antiseptic 
cotton and bandages. 

If any acid gets upon the clothing, apply the ammonium hydroxide solution 
from your work-desk at once. 

What to Aim For. -- The excellence of your laboratory work will be judged 
by the correctness and completeness of your observations; by the accuracy of 
your deductions; by the quality and yield of the substances you prepare; by 
the completeness and accuracy of your reports; and by the skill, rapidity, and 
neatness which you display in the laboratory. 



PART I 
EXPERIMENTS IN GENERAL CHEMISTRY. 

Experiment I. 

To Determine the Proportion by Weight in which Oxygen and Copper 
Combine to Form Cuprous Oxide. 

Assigned Reading. Smith, pp. 41-52. 

Discussion. In general this could be done either by synthesis or analysis; 
that is, either the weight of oxygen required to convert one gram of copper into 
cuprous oxide, or the weight of copper left by a known weight of cuprous oxide 
after the complete removal of the oxygen could be determined. However, when 
copper is oxidized by heating in the air, unless special precautions are taken, a 
mixture of cuprous and cupric oxides is obtained. Analysis is, therefore, the more 
satisfactory method in this case ; and the most convenient means for the removal 
of the oxygen, that is, the most convenient " reducing agent," is ordinary illumi- 
nating gas. The cuprous oxide must, however, be hot, else the gas will not 
act upon it. 

Directions. Fit a one-hole rubber stopper in each end of a hard-glass tube, 
and attach the tube to the ring stand by means of a clamp, keeping the clamp 
very close to one end of the tube. Into one of the rubber stoppers in the hard- 
glass tube fit a U-tube provided with a cotton-wool plug at its exit end, and con- 
taining granular soda-lime (a mixture of lime and sodium hydroxide, two very 
hygroscopic substances), and fit into the rubber stopper in the other end of the 
tube a short elbow of small glass tubing : with its arm pointing upward. 

Obtain some powdered cuprous 2 oxide from the instructor. Weigh a clean, 
dry, porcelain boat, spread about one gram of the cuprous oxide along its bottom, 
and weigh again. Place the boat with its contents in the middle of the hard-glass 
tube, and connect the free end of the U-tube by means of rubber tubing with the 
laboratory supply of illuminating gas. Fix the hard-glass tube in a slightly 
inclined position, in such a way that the gas enters at the upper end(?). At 
this point secure the approval of the instructor for your apparatus as set up. 

1 The cut ends of all glass tubing should invariably be smoothed with a file, or far better should be 
" fire-polished," that is, held long enough in a Bunsen flame to soften, and so round over, the sharp edges. 

2 Most commercial cuprous oxide contains a large percentage of cupric oxide. Pure cuprous 
oxide may conveniently be prepared by boiling pure cuprous chloride with a solution of potassium 
hydroxide, washing thoroughly by decantation (see footnote 1, page 109) with hot water, followed by 
alcohol and ether. 

7 

\ 



LABORATORY MANUAL OF GENERAL CHEMISTRY 


Pass a gentle stream of illuminating gas through the tube for two minutes 

and then (?) light the gas issuing from the glass elbow. The flame should not be 
more than half an inch long. Heat the boat, using the wing-top, at first gently, 
but afterwards, when the cuprous oxide is apparently all reduced, as strongly as 
your Bunsen burner will permit. Allow the tube to cool very slowly but com- 
pletely; then shut off the gas supply, remove the boat, and weigh again. Keep 
watch for any substance accumulating in the cooler portions of the tube. Note 
any change which may have occurred in the hard-glass tube itself. Carefully 
preserve the boat and contents for the next experiment. 

Calculate the weight of oxygen which was combined with one gram of copper. 

Outside Questions. 

(1) What substance did you observe accumulating in the cooler portions of 
the tube ? Tell what its source must have been. What other products 
may have been formed (R 493 and 513) ? 

(2) How does cuprous oxide differ from cupric oxide in appearance and in 
chemical properties (R 622) ? 

(3) How may pure cupric oxide be prepared without any danger of having 
cuprous oxide present as an impurity ? 

(4) If any change in the hard-glass tube was observed explain its cause 
(R607). 



Experiment I. 

To Determine the Proportion by Weight in which Oxygen and Copper 
Combine to Form Cuprous Oxide. 

LABORATORY NOTES. 

Name Section 

Date Desk No. 



Experiment II. 

To Determine the Proportion by Weight in which Sulphur and Copper 
Combine to Form Cuprous Sulphide. 

Assigned Reading. Smith, pp. 40-52. 

Discussion. In contrast to cuprous oxide, cuprous sulphide can easily be 
made by heating a mixture of the two constituent elements (?R). It is more 
convenient, therefore, in this case to follow the method of synthesis rather than of 
analysis. For a given quantity of copper an excess of sulphur must be used, and 
this excess distilled off after the chemical combination has been effected. During 
this process the hot cuprous sulphide must not be allowed to come in contact 
with air (?). Some cupric sulphide escapes decomposition in the crucible 
because of the presence of sulphur vapor. The heating must, therefore, be con- 
tinued in a current of some indifferent gas, that is, of some gas which does not 
act upon cuprous sulphide. Illuminating gas is a convenient one for this purpose. 

Directions. Transfer the reduced copper from Experiment I carefully and 
completely to a clean, dry, porcelain crucible resting on a sheet of clean, white 
paper. The last traces can be transfered easily by means of the small camel's- 
hair brush. Preserve the boat carefully, as it must be used later in the experi- 
ment. Add about an equal amount of roll sulphur to the crucible, put its cover 
on, and place it upon a pipe-stem triangle supported on a tripod. Heat slowly 
with your Bunsen burner until no more flames of burning sulphur can be seen 
escaping from under the edge of the cover, increasing the heat slowly up to the 
full capacity of the burner. Do not remove the cover while the crucible is hot. 
Allow the crucible to cool, place it, as before, on the white paper, then remove 
the cover and pulverize the contents with a small test-tube or rounded glass rod, 
being careful not to lose any of the powder. Transfer the powder back to the 
same porcelain boat and heat this for several minutes to redness in a current of 
illuminating gas, using the same apparatus and procedure as in Experiment I. 
Allow the apparatus to cool, and again weigh the boat. 

Calculate the weight of sulphur combined with one gram of copper. 

Outside Questions. 

(1) From the result of this experiment and that of Experiment I calculate the 
weight of sulphur which should combine with sixteen grams of oxygen. 

(2) Why was it necessary to keep the crucible covered during the heating ? 

(3) In addition to its purely protective action the illuminating gas produced 
another favorable effect. What was it (R 513 and 625) ? 

(4) What extra weighing in this experiment could you have made to prove 
that the cupric sulphide was not wholly decomposed in the crucible ? 

(5) How can cupric sulphide be obtained (R 625) ? 



Experiment II. 

To Determine the Proportion by Weight in which Sulphur and Copper 
Combine to Form Cuprous Sulphide. 

LABORATORY NOTES. 

Name Section 

Date Desk No. . 



Experiment III. 



To Determine the Proportion by Weight in which Oxygen and Sulphur 
Combine to Form Sulphur Dioxide. 

Assigned Reading. Smith, pp. 41-52. 

Discussion. Here again synthesis is the more convenient method; but since 
the reaction product (S0 2 ) is gaseous, it must be collected by some suitable absorb- 
ent (soda-lime) in order to be weighed conveniently. 

Directions. Set up the same hard-glass tube as was used in Experiments 
I and II (see Fig. 1). Prepare a second U-tube identical with the one used in 




Aspirator 



Soda-lime 




Fig. 1. 

Experiment I ; provide it with a suspension made of copper wire for hanging it to 
the balance arm, and attach it by means of a one-hole stopper to one end of the 
hard-glass tube. Attach an aspirator to the other end of this U-tube. A suitable 

11 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

aspirator for this purpose can be constructed by fitting a glass elbow and a glass 
siphon tube into the neck ,of your large bottle, using a two-hole rubber stopper. 
The flow of water from the aspirator, and hence the flow of air through the hard- 
glass tube, can be regulated by attaching a short piece of rubber tubing, provided 
with a screw pinchcock, to the lower end of the siphon tube. 

To the free end of the hard-glass tube attach by means of rubber tubing a 
second U-tube for use. Vith liquids. Into the open end of this U-tube pour just 
enough con. H 2 S0 4 to seal the bend. Have ready also a second Bunsen burner. 
Weigh out 0.15-0.25 gram of roll sulphur in a clean boat and place this in the 
hard-glass tube with its end 4-5 cm. from the rubber stopper where the air enters. 
In order to dry the tube thoroughly set the aspirator running rapidly enough to 
produce two or three bubbles of ah- per second through the H 2 S0 4 , and gently 
warm the hard-glass tube through its whole length. After four or five minutes 
stop the aspirator, disconnect the second soda-lime U-tube, weigh, and re-connect 
it. (?) Now place a second Bunsen burner midway between the boat and the 
exit end of the tube, and turn on a strong flame so that that portion of the tube 
becomes very hot. This is to prevent any sulphur vapor escaping unburnt from 
the tube. Then start the aspirator at a slightly faster rate, and heat the sulphur 
gently at the end nearer the air supply until it catches fire. Discontinue the 
heating at once and resume only if the flame threatens to go out, and at the 
end to make sure that all of the sulphur has burned. The reason for this caution 
hi heating is to avoid, so far as possible, the distillation of sulphur out of the 
boat. Allow the aspirator to run for five minutes after the last of the sulphur has 
disappeared, then disconnect the weighed U-tube and weigh again. From the 
increase, calculate the weight of sulphur combined with one gram of oxygen. 
Estimate roughly, from the volume of your aspirator and the number of times 
you had to refill it, how much air was sucked through the tube while the sulphur 
was burning. 

If, for any reason, it should be necessary to repeat this experiment, be sure 
to refill the weighed U-tube with fresh soda-lime. 

Outside Questions. 

(1) Calculate the weight of sulphur combined with sixteen grams of oxygen. 
Compare this result with that of Experiment II. What two laws does this 
demonstrate ? 

(2) Compute approximately what percentage of the oxygen in the air sucked 
through the tube was consumed. 

(3) Compute approximately how great an error in your results in this experi- 
ment must have been caused by the carbon dioxide in the air, assuming 
that there was one- tenth of one percent by weight of this substance in 
the air of the laboratory. 

(4) Write equations for the absorption of the sulphur dioxide and the carbon 
dioxide, as well as for the combustion of the sulphur. 

12 



Experiment III. 

To Determine the Proportion by Weight in which Oxygen and Sulphur 
Combine to Form Sulphur Dioxide. 

LABORATORY NOTES. 
Name Section 

s 

Date ... Desk No. 



Experiment IV. 
Oxidation and Reduction. 

Discussion. Oxidation, in the simplest case, means the chemical combi- 
nation of oxygen with a substance; reduction is the opposite of oxidation, and 
hence means the removal of oxygen from a substance. The substance which 
furnishes oxygen is the oxidizing agent that which removes it is the reducing 
agent. Thus, when carbon burns in the air forming carbon cjioxide, gaseous 
oxygen is the oxidizing agent, and carbon the reducing agent. When carbon and 
iron oxide are heated together and carbon dioxide and metallic iron are formed, 

C + 2FeO = CO 2 + 2Fe, 

the oxide of iron is the oxidizing agent, and carbon is the reducing agent. The 
process may be looked upon either as an oxidation of the carbon , or as a reduc- 
tion of the iron oxide. 

Oxidation, however, is not restricted simply to the addition of oxygen to a 
molecule as a whole. Oxygen may combine with one or more atoms in the 
molecule and tear it, or them, out of the molecule. Thus, when hydrogen sul- 
phide is burned with an inadequate supply of air, sulphur and water are formed; 

2H 2 S + O 2 = 2H 2 + 2S. 

The hydrogen sulphide is oxidized by the removal of hydrogen. 

Reduction, similarly, is not restricted to the removal of oxygen. It may 
also signify the addition of hydrogen, or some substance which combines readily 
with oxygen, to the molecule. Thus when sulphur is heated in hydrogen, hydro- 
gen sulphide is formed, 

S -f- H 2 = H 2 S, 

and the sulphur is said to be reduced to hydrogen sulphide, although no oxygen 
has been removed. 

Finally, the terms oxidation and reduction are applied to certain reactions 
in which no oxygen or hydrogen is involved. This usage is justified by the fact 
that the same products can be obtained, perhaps less directly, by the employ- 
ment of oxygen or hydrogen. 

Thus, when stannous chloride and ferric chloride react to form stannic 
chloride and ferrous chloride, 

SnCl 2 + 2FeCl 3 = SnCl 4 + 2FeCl 2 , 

the stannous chloride is said to be oxidized and the ferric chloride reduced, 
although no oxygen or hydrogen is directly involved. This is justified, because 

13 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

stannic chloride can also be produced by oxidizing stannous chloride to stannic 
oxy chloride with oxygen and treating the latter with hydrochloric acid; 

2SnCl 2 + O 2 = 2SnCl 2 O, 
SnCl 2 + 2HC1 = SnCl 4 + H 2 0. 

Similarly, the ferric chloride can be reduced to ferrous chloride by heating with 
hydrogen under proper conditions ; 

2FeCl 3 + H 2 = 2FeCl 2 + 2HC1. 

Later it will be seen that all of these different types of oxidation and reduc- 
tion can be grouped under one head, and involve fundamentally the same 
phenomenon. 

Directions. 

(a) Dilute a few drops of ferrous sulphate solution with 5 c. cm. water in a 
test-tube and add a few drops of potassium ferrocyanide solution (Ei). 
Repeat this test using ferric chloride solution instead of ferrous sulphate 
solution (E 2 ). 

(6) Dilute a few drops of ferrous sulphate solution as before, neutralize the 
acid present with ammonium hydroxide solution, warm gently, and then 
aerate the solution by closing the test-tube with your thumb and shaking 
vigorously for a few moments (E 3 ). . Acidify the solution with a little 
dilute sulphuric acid (E 4 ), and test for ferric iron as above. 

(c) Dilute a little of the ferrous sulphate solution as before, add a few drops 
of chlorine water (E 5 ), and again test for ferric iron. 

(d) Dilute one cubic centimeter of silver nitrate solution with about four 
cubic centimeters of water. Add a cubic centimeter of ferrous sulphate 
solution and heat (E 6 ). 

Outside Questions. 

(1) Does the reaction represented by (E 4 ) involve an oxidation or a reduction ? 

(2) Write the equation for the oxidation of ferrous sulphate in the presence 
of sulphuric acid. 

(3) Name the oxidizing and reducing agents in (c) and (d). 

(4) Explain how the same substances could be made, using oxygen as an 
oxidizing agent and hydrogen as a reducing agent. 



14 



Experiment IV. 
Oxidation and Reduction. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment V. 

Hydrogen Peroxide. Peroxides. 
Assigned Reading. Smith, pp. 302-309. 

Directions. 

(a) Weigh out 2 grams of sodium peroxide and add it, a very little at a time, 
to 100 c. cm. of ice- water. Then add dilute sulphuric acid to the solution, 
a few drops at a time, until the solution is distinctly acid to litmus paper 
(Ei). What does the solution now contain ? 

(6) Transfer a portion of this solution to a test-tube, and add finely powdered 
manganese dioxide. Determine what the escaping gas is (E 2 ). Does 
trie manganese dioxide appear to have undergone any alteration ? 

(c) Dilute 10 c. cm. of this solution ten times, and to a portion of the diluted 
solution add a few drops of starch-iodide solution (i. e., a solution con- 
taining starch emulsion and potassium iodide) (E 3 ). (In writing the 
equation for this reaction remember that the prepared solution contained 
an excess of sulphuric acid.) To what class of substances does this show 
hydrogen peroxide to belong ? 

(d) To another portion of the diluted solution add 3 c. cm. of ether, shake, 
and then add a single drop of a solution of potassium dichromate, and 
again shake. 

(e) To a portion of the original solution add a little potassium permanganate 
solution. (?) 

(/) Shake some lead dioxide with ice- water, and add dilute sulphuric acid as 
in the case of the sodium peroxide. Filter and test the filtrate for hydro- 
gen peroxide as in (d). Are all higher oxides peroxides ? 

Outside Questions. 

(1) (A) In preparing the solution of hydrogen peroxide why was ice-water 
employed ? 

(B) How could you prove definitely whether the manganese dioxide had 
undergone any alteration ? 

(2) (A) What is an action such as that exerted by the manganese dioxide 
called ? 

(B) Name other instances of the same sort. 

(3) Is hydrogen peroxide the only substance which will produce the results 
observed in (c) ? in (d) (R 729) ? 

15 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(4) (A) What general name can be given for the behavior of hydrogen per- 
oxide in (e) ? 

(B) How does this agree with the results of (c) ? 

(C) Remembering how difficult it is to break up oxygen gas into its atoms 
by heat, how do you explain the behavior of hydrogen peroxide in (e) 
(R 307) ? 

(D) Write the equation for this reaction. 

(5) On the basis of (e) how would you distinguish between a peroxide and a 
dioxide (R 308) ? 



16 



Experiment V. 
Hydrogen Peroxide. Peroxides. 

LABORATORY NOTES. 

Name Section 

Date ... Desk No. 



Experiment VI. 
Autoxidation. 

Discussion. When substances combine with gaseous oxygen the first 
product appears in general to be a simple compound formed by the direct addi- 
tion of one molecule of the gas to one molecule of the substance in question. 
These simple addition compounds are strong oxidizing agents and are usually 
very unstable. To prevent their decomposition the temperature must be kept 
low. 

Thus, when hydrogen burns in liquid oxygen, hydrogen peroxide is formed 
and is then immediately frozen by the extreme cold; 

H 2 + O 2 = H 2 2 . 

When hydrogen burns in oxygen at ordinary temperatures doubtless the same 
reaction occurs first, but the hydrogen peroxide is decomposed at the high 
temperature of the flame into oxygen and the more stable water; 

2H 2 2 = 2H 2 + 2 . 

Similarly, when metals are oxidized by the air at ordinary temperatures 
without burning, direct combination of the atoms of the metals with the mole- 
cules of gaseous oxygen occurs and peroxides of the metals result. When these 
peroxides are treated with dilute acids hydrogen peroxide is of course formed, 
and its presence can be demonstrated easily by appropriate tests. This case is 
illustrated in the present experiment. 

These peroxides oxidize many substances which gaseous oxygen does not 
affect. We have then the rather surprising phenomenon of a reducing agent, the 
metal, combining with an oxidizing agent, the oxygen, to produce a still more 
active oxidizing agent. One would certainly expect in advance that when part 
of the chemical affinity of oxygen had been utilized in combining with the metal, 
there would be less tendency for it to combine with other reducing agents. Part 
of the explanation of this phenomenon doubtless lies in the fact that the oxygen 
atoms are held very firmly together in the molecules of oxygen gas. 

An oxidation such as this, where by the oxidation of a reducing agent a more 
active oxidizing agent is produced, is called autoxidation. 

Directions. Amalgamate a piece of lead foil weighing between a quarter 
and a half a gram by treating it in a test-tube first with dilute nitric acid (Ei), 
and then with a solution of mercurous nitrate (E 2 ). Cleanse the lead by repeated 
rinsing in the test-tube with water, and then pour 1 c. cm. of mercury upon it. 
Prepare 50 c. cm. of 0.2 M sulphuric acid in an Erlenmeyer flask, add the amal- 
gamated lead, and shake vigorously at frequent intervals for ten minutes (E 3 ). 

17 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Remove portions of the liquid from time to time and test for hydrogen 
peroxide by means of starch-iodide solution. Allow the test solutions to stand 
for some time, or add a few drops of very dilute ferrous sulphate solution which, 
acting as a catalyst, greatly hastens the oxidation of the hydriodic acid and hence 
the development of the iodine-starch color. 

At the conclusion of the experiment, pour the mercury into the bottle 
labeled " mercury residues." 

Outside Questions. 

(1) To what was the turbidity of the above solution due (R 704) ? 

(2) What was the function of the mercury ? 



18 



Experiment VI. 
Autoxidation. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment VII. Solutions. Gases in Liquids. 
To Determine the Solubility of Air in Water. 
Assigned Reading. Smith, pp. 153-155. 

Discussion. Gases which are difficult to liquify are almost invariably but 
slightly soluble in liquids. Air is no exception to this rule. 

The solubility of air in water is conveniently measured by sweeping the air 
out of its saturated solution by means of water vapor, that is, by boiling the 
solution and collecting the air after the water vapor has been condensed. 

All of the water vapor is not condensed because, of course, water has a 
considerable vapor pressure even at room temperature. It is evidently necessary 
to allow for this in computing the amount of air dissolved. 

Directions. Half fill your 500 c. cm. flask with water, close it with a 
stopper, or your hand, and then shake violently for a minute or two until the 
water is thoroughly aerated. Read the temperature of the water and the height 
and temperature of the barometer. 1 Then fit a one-hole stopper and a delivery 
tube to your 100 c. cm. flask, and measure the volume of the flask up to the 
lower surface of the stopper. Fill the flask and delivery tube completely with 
the aerated water, clamp to your retort stand, and boil, collecting the air in a 
small test-tube inverted in a vessel of cold water. When no more air comes over 
with the steam, first submerge the whole test-tube in the water, holding the 
test-tube by its lower end, so that the air may attain the temperature of the 
water, then raise the test-tube sufficiently to equalize the levels of the water 
inside and outside, and mark the level either with a file or by using a thin rubber 
ring cut from a piece of rubber tubing. Measure the volume of the test-tube to 
this mark, and read the temperature of the water in which the test-tube was 
submerged. 

1 Because of the unequal thermal expansion of the mercury as compared with the scale of the 
barometer, the following corrections, in millimeters, must be subtracted from observed barometric 
readings between 740 and 770 mm. at the specified room temperatures in order to obtain the height to 
which the mercury would rise were the barometer cooled to C. 

BAROMETER CORRECTIONS 

Temp. Correction Temp. Correction 

mm. mm. 

12 1.6 22 2.9 

14 1.8 24 3.2 

16 2.1 26 3.4 

18 2.3 28 3.7 

20 2.6 30 4.0 

19 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Outside Questions. 

(1) Calculate from your observed data the solubility of air in water, ex- 
pressed in cubic centimeters of dry air l (R 88) at the observed pressure 
and temperature, per liter of water. On this basis what volume of air 
would be dissolved at 50 cm. pressure if the volume were measured at 
this same pressure (R 154) ? 

(2) Does the dissolved air have the same composition as ordinary air ? 
why (R 416) ? 

1 VAPOR PRESSURE OF WATER 

(Aqueous Tension) 
In Millimeters of Mercury 

Temp. Pressure Temp. Pressure Temp. Pressure 

10 9.2 20 17.5 40 55.1 

11 9.8 21 18.6 50 92.3 

12 10.5 22 19.8 60 149.2 

13 11.2 23 21.0 70 233.5 

14 11.9 24 22.3 80 355.0 

15 12.7 25 23.7 90 526.0 

16 13.6 26 25.2 100 760.0 

17 14.5 27 26.7 110 1075.0 

18 15.4 28 28.3 120 1489.0 

19 16.4 29 30.0 130 2026.0 

30 31.8 140 2710.0 



20 



Experiment VII. 
To Determine the Solubility of Air in Water. 

LABORATORY NOTES. 

Name Section 

Date ... Desk No. .. 



Experiment VIII. Solutions. Solids in Liquids. 
To Determine the Solubility of a Solid in a Liquid. 

Assigned Reading. Smith, pp. 156-161. 

Discussion. At the surface of contact between a solid substance and its 
saturated solution a condition of equilibrium evidently exists (R 152-153). 
That is, the tendency for the molecules of the solid to dissolve is just balanced 
by the back-pressure, or osmotic pressure of the molecules which have already 
gone into solution. It takes some time for this equilibrium to be precisely 
established, for the molecules of the solid must dissolve and be uniformly dis- 
tributed throughout the solution, and the more nearly equilibrium is attained, 
the slower this process becomes. Evidently the larger the surface of contact the 
more rapidly this adjustment will proceed, so the following directions specify a 
considerable excess of the finely powdered solute. 

Directions. Grind 6 grams of potassium dichromate to a fine powder, and 
shake it at intervals for ten minutes with 25 c. cm. of water in your 100 c. cm. 
flask. Meanwhile clean and weigh accurately a porcelain crucible. Take the 
final temperature of the solution. Obtain a 10 c. cm. pipet on temporary order 
from the store-room. Using it, transfer exactly 10 c. cm. of the clear solution 
to the weighed crucible, weigh again, and then evaporate to dryness on the steam 
bath, and heat gently for a little while on the hot plate. Finally weigh the 
dry crucible and its contents. Calculate the weight of potassium dichromate 
in 100 grams of solution. Return the pipet to the store-room. 



Experiment IX. Solutions. Solids in Liquids, continued. 

To Observe the Effect of Temperature on Solubility. 

Directions. 

(a) Boil 6 grams of sodium chloride with 10 c. cm. of water in a test-tube. 
Pour the clear liquid immediately into another test-tube and cool this 
latter under the tap. Is this salt much more soluble in hot than in 
cold water ? 

(6) Boil 6 grams of potassium dichromate in 10 c. cm. of water in a test-tube. 
Cool the test-tube under the tap. Explain the result. 

21 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Outside Questions. 

(1) 100 grams of water will- dissolve 4.64, 45.4, and 108.2 grams of dichro- 
mate at 0, 60, and 105. From this data, from that which you secured 
in Experiment VIII, and from (R 157), draw the solubility curves of 
sodium chloride and potassium dichromate on a single diagram. Read 
off from these curves the solubilities of sodium chloride and of potassium 
dichromate at 15 and at 100. 

(2) Knowing the specific gravity of a saturated solution of potassium dichro- 
mate at 15 to be 1.062, calculate the molar solubility, that is, the molar 
concentration of the saturated solution, at that temperature. 



Experiment VIII. 
To Determine the Solubility of a Solid in a Liquid. 

LABORATORY NOTES. 

Name Section 

Date Desk No. . 



Experiment IX. 
To Observe the Effect of Temperature on Solubility. 

LABORATORY NOTES. 

Name Section 

Date ... Desk No 



Experiment X. Solutions. 
Supersaturated Solutions. 

Discussion. Water can be cooled a little below its true freezing point 
without freezing, provided no crystals of ice are present and the water is not 
jarred. It can also be heated above its true boiling point without boiling, pro- 
vided it is free from dissolved gases. Other liquids can be "supercooled " and 
" superheated " in a similar fashion. 

In the same way, when a saturated solution of a substance the solubility 
of which decreases with falling temperature is cooled, crystals of the dissolved 
substance do not at once appear, provided the solution is entirely free from 
them at the start. 

Directions. Place 10 grams of crystalline sodium acetate in a 100 c. cm. 
flask; moisten with a little water, and heat gradually over a small flame until 
the salt is all dissolved and steam issues freely from the mouth of the flask. 
Allow the flask to cool somewhat in the air, and then cool thoroughly under the 
tap. When cold, drop into the clear liquid a small crystal of sodium chloride 
and then the smallest detectable fragment of sodium acetate (?). This cycle 
may be repeated indefinitely if the water which evaporates is replaced. 

Outside Questions. 

(1) Prove from your observations in this experiment what temperature 
change must occur when sodium acetate dissolves in water. 

(2) How would you prepare a supersaturated solution of a substance whose 
solubility decreases with rising temperature ? 



Experiment XI. Solutions. 

To Observe the Effect of Dissolved Solids on the Vapor Pressure. 
Assigned Reading. Smith, pp. 161-162. 

Discussion. If an efficient air-pump is at hand the vapor of a liquid may be 
simply measured by first pumping out the air in a closed vessel containing the 
liquid and then connecting this vessel with an arm of a U-tube containing mer- 
cury, the other arm being open to the air. The difference in level between the 
mercury in the two arms of the U-tube evidently represents the difference be- 
tween the vapor pressure of the liquid and the atmospheric pressure. Subtract- 
ing this difference from the barometer reading would then give the vapor pressure 

23 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

of the liquid. To find the effect on the vapor pressure of dissolving a solid in a 
liquid, one would first measure the vapor pressure of the pure liquid in this way 
and then, after dissolving a weighed quantity of the solute in the liquid, repeat 
the same measurement on the resulting solution. 

It is simpler to adopt a differential method; that is, to measure the differ- 
ence between the vapor pressure of the pure liquid on the one hand, and that of 




Fig. 2. 



the solution on the other. This is the method adopted in the following experiment. 
It is carried out by connecting a pressure gauge between two closed test-tubes, 
each containing some of the pure solvent. When temperature differences have 
been eliminated, the pressure gauge will show no difference of pressure between 
the two test-tubes. A weighed quantity of a solid substance is then dissolved in 
the solvent in one of the closed test-tubes. The effect of this addition on the 
vapor pressure of the solvent is measured by the pressure gauge. 



24 



EXPERIMENT XI 

Directions. 

(a) Construction of Apparatus. 

Construct the apparatus as illustrated in Fig. 2. It consists of two large 
test-tubes, A and B (160 X 25 mm.), 1 each tightly closed with a two- 
hole rubber stopper (No.*5). Construct a pressure gauge, or manometer 
(U), consisting of a U-tube containing water, with its two limbs 30 cm. 
long, and with each of its ends bent downward and fitted into one of the 
rubber stoppers. Make one of these ends long enough to reach to within 
4-5 cm. of the bottom of the test-tube. To make the manometer less 
fragile groove a cork and wire it in place between the upper ends of the 
manometer limbs. For this same purpose the test-tubes may advan- 
tageously be bound together with a few elastic bands. Fit two exit tubes 
9-10 cm. long into the remaining holes in the rubber stoppers, and close 
them at the tops with short pieces of rubber tubing and pinchcocks. 

The arrangement for dropping the solid substance into the solvent in one 
of the test-tubes (B) consists of a small (100 X 10 mm.) test-tube, with 
either an enlargement or a groove blown on its closed end. By means 
of this device a wire or string can be attached to the test-tube, and 
the latter thereby hung, up-side-down, from the end of the exit tube. 
Close the small test-tube by a cork, attaching the latter firmly to the end 
of the manometer tube by means of small elastic bands. The cork can 
then be withdrawn from the test-tube simply by pulling the exit tube up 
through the rubber stopper. Sink the two large test-tubes as deeply as 
possible into a tall beaker (550 c. cm.) filled with water. Use a retort 
stand and clamp, if necessary, to accomplish this. 

(6) Manipulation. 

Fill the manometer half -full of water. Measure 5 c. cm. of acetone into 
each of the large test-tubes. Dry the small test-tube and fill it with 
benzoic acid, weighing the acid with an accuracy of 2-3%. Insert 
the cork stopper and fasten it and the test-tube in place. Stir the 
water in the beaker, close both exit tubes, and let the whole apparatus 
stand undisturbed for at least ten minutes. Equalize the pressure in both 
tubes by opening the exit tube momentarily. After five minutes, if no 
change in the manometer limbs has occurred, all temperature differences 
evidently will have been equalized. Now pull the small test-tube clear of 
the cork stopper, open the exit tubes momentarily to restore the equality 
of pressure disturbed by pulling out the small tube. Then shake the 
test-tube vigorously so as to wash and shake the benzoic acid into the 
acetone. After solution is complete, let stand for fifteen minutes,' with 

1 If more convenient, one of the test-tubes (A) may be smaller (160 X 20 mm.), and closed with a 
smaller rubber stopper (No. 2). 

25 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

occasional shaking, but without removing the test-tubes from the beaker, 
then read the difference of levels in the manometer with a millimeter 
rule. Knowing that the density of mercury is 13.6, compute the change 
in vapor pressure of acetone in millimeters of mercury per gram of benzoic 
acid, per liter of acetone. Suggest an explanation of the large initial 
pressure change. 

Outside Questions. 

(1) Compute from your results the change in the vapor pressure of acetone 
per gram molecule (A) per liter, and (B) per 1000 grams of acetone. The 
formula of benzoic acid is C 6 H 5 COOH. The density of acetone is 0.795 
(R 503). 

(2) Knowing that the vapor pressure of acetone increases by 10 mm. of 
mercury per degree at room temperature, compute how great a difference 
in temperature between the contents of the two test-tubes would produce 
a difference in level of one millimeter in the manometer. 



26 



Experiment X. 
Supersaturated Solutions. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XI. 
To Observe the Effect of Dissolved Solids on the Vapor Pressure. 

LABORATORY NOTES. 

Name Section .. 

Date ... Desk No 



Experiment XII. Solutions. 

To Observe the Effect of Dissolved Solids on the Boiling Point. 
Assigned Reading. Smith, pp. 115-118 and 161-162. 

Discussion. Since the boiling point of a liquid is merely the temperature at 
which the vapor pressure of the liquid becomes equal to that of the surrounding 
atmosphere, and since Experiment XI demonstrates that the vapor pressure of a 
liquid is altered by dissolving a solid substance in it, the conclusion follows that 
the boiling point of a solution must differ from that of the pure solvent. The 
following experiment is designed to measure this difference in a particular case. 

Directions. Slip a one-hole rubber stopper over the upper end of a ther- 
mometer, and by attaching a small clamp to this stopper, support the thermometer 
in an upright position on a retort stand. Support a medium-sized test-tube con- 
taining 10 c. cm. of water by means of a similar clamp attached lower down to 
the retort stand in such a way that the bulb of the thermometer is just immersed 
in the water and is in the exact axis of the test-tube. Boil the water, using a 
small Bunsen flame, and read the temperature. Afterwards read the barometer. 
Now add 2 grams of dry calciUm chloride to the water, and after it has all dis- 
solved, determine the boiling point of the solution. Replace any water which 
may have boiled away, and add an additional 4 grams of calcium chloride, and 
repeat the determination. 

Outside Questions. 

(1) Calculate from the data obtained in this experiment the change in boiling 
point per gram molecule of calcium chloride per liter of water. 

(2) Plot on co-ordinate paper the vapor pressures of water at each 10 interval 
between 10 and 100, using the values given on page 20, and draw a curve 
connecting these points. Also indicate on this same diagram the points 
you have determined in this experiment on the vapor pressure curves of 
the two calcium chloride solutions. Draw in these two curves, assuming 
that they are parallel to the vapor pressure curve of water. According to 
your curves, what are the vapor pressures of the two calcium chloride 
solutions- at 100? 



27 



Experiment XII. 
To Observe the Effect of Dissolved Solids on the Boiling Point. 

LABORATORY NOTES. 

Name Section 

Date ... Desk No 



Experiment XIII. 

The Velocity of Chemical Reaction. 

Discussion. When a substance decomposes spontaneously, a molecule at 
a time, it is evident tha.t the reason for the decomposition must lie in some condi- 
tion of instability set up inside the molecule. Not all the molecules get into this 
unstable condition at once, or decomposition would be instantaneous and com- 
plete. Moreover, the number which do so will be affected by changes in tempera- 
ture. A rise in temperature, for instance, will involve a greater frequency and 
violence of molecular collisions, and this will favor the attainment of unstable 
conditions inside the molecules. At constant temperature, however, the fraction 
of the molecules present which in a given interval of time become unstable and 
hence decompose, will be strictly constant. The number of grams of the original 
substance which decompose per second will then be proportional simply to the 
number of grams of the original substance present. In other words, in any given 
volume, the velocity of this reaction will be proportional to the concentration of 
the reacting substance. 

This conclusion may be stated algebraically by writing 

V = kC, 

where V represents the velocity of the reaction, C the concentration of the react- 
ing substance, and k a constant, the so-called velocity constant of the reaction. 
This last term will, of course, have different values for difference reactions, 
and also for the same reaction at different temperatures. For rapid reactions 
k will be large; for slow ones, small. To illustrate this, suppose, for a particular 
reaction the value of k is 0.001, where the velocity is expressed in gram molecules 
decomposed per liter, per second. Then, at unit concentration, that is, at a 
concentration of one gram molecule per liter, the velocity of the reaction would be 

V = 0.001 X 1 = 0.001 g. mole, per sec. 

When, however, the chemical reaction is not merely the decomposition of 
a single molecule, but requires the combination of two or more molecules for its 
consummation, the velocity of the reaction is proportional to the concentration 
of each of the reacting molecules. This follows from the fact that, at constant 
temperature, the probability of the occurrence of one of the necessary collisions 
of the different molecules is proportional to the concentration of each of the 
reacting molecules. It further follows that the velocity of the reaction, since it is 
proportional to each of the concentrations, is also proportional to the product of 

29 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

these concentrations. Calling these concentrations of the reacting molecules 
C a , Cb, C c , etc., the algebraic expression for the velocity of reaction then is: 

V = kC a X C b X C c etc. 

Applying this expression, for the sake of illustration, to the case of hydrogen 
and iodine, which combine rapidly at a slightly elevated temperature to form 
hydrogen iodide according to the equation, 

H 2 + I 2 = 2HI, 
the velocity of the reaction would be 

V = kC H2 X Ci 2 . 

That is, at constant temperature, the number of grams of hydrogen iodide formed 
per second in a given volume will be directly proportional to the product of the 
concentrations of the hydrogen and the iodine. 

In the following experiment, where a mixture of ferric chloride and ferric 
thiocyanate is slowly reduced by staijinous chloride, the progress of the reaction is 
followed by observing the slow bleaching of the solution. By counting the seconds 
required to reach a certain standard tint, the velocities at different concentrations 
may roughly be compared. 

Directions. 

(a) Make a series of transverse file scratches at 2.5 cm. intervals along a 20 cm. 
length of glass tubing, 5 mm. in bore. The volume of the tube between 
scratches will be very nearly one-half a cubic centimeter. 

(6) To 200 c. cm. of water in a flask add 2 c. cm. of ferric chloride solution, 
using the above tube as a pipet. Then add 5 c. cm. of ammonium thiocya- 
nate solution (Ei), followed by 10 c. cm. of concentrated hydrochloric acid. 
Mix thoroughly by shaking. 

(c) Place 15 c. cm. of this solution in each of several medium-sized test-tubes. 
Prepare a comparison tube by diluting 5 c. cm. of the solution with 
10 c. cm. of water. Label this tube. Rinse out the above pipet and with 
it add one-half a cubic centimeter of stannous chloride solution to one of 
the 15 c. cm. portions of the original solution (E 2 ). The instant the 
stannous chloride solution has been added, close the test-tube with your 
thumb, shake and note the exact time of the addition as indicated by the 
second-hand of your watch. Hold this tube and the comparison tube 
close together and observe them against a uniform, brightly illuminated 
background. Note the second on your watch when the reacting solution 
has attained the exact shade of the comparison solution. Repeat this 
experiment with a fresh portion of solution, but add a whole cubic centi- 
meter of stannous chloride solution. Repeat both of these experiments 
until you are satisfied with the accuracy of your results. Finally, dilute 

30 



Experiment XIII. 
The Velocity of Chemical Reaction. 

LABORATORY NOTES. 

Name Section .. 

Date Desk No - 



EXPERIMENT XIII 

75 c. cm. of the original solution with an equal volume of water; dilute 
5 c. cm. of this solution with 10 c. cm. of water for a comparison solution, 
and add to successive 15 c. cm. portions, as before, one-half, and again one 
cubic centimeter quantities of the stannous chloride solution. Repeat 
your measurements at one of the concentrations, but warm each solution 
10 before mixing. Record your results in tabular form. Compute the 
average of your results with each of the four concentrations. What do 
you conclude is the effect of changes in concentration and in temperature 
on the velocity of this reaction ? 

Outside Questions. 

(1) Do your measurements give the true initial velocity of this reaction ? If 
not, what velocity do they measure ? 

(2) Assuming that the stannous chloride in this experiment reacted merely 
with the ferric chloride, write the algebraic expression showing how the 
velocity of this reaction will vary with the concentrations of the reacting 
substances. 

(3) On the basis of (2), what would be the effect on the initial velocity of 
doubling the concentrations (A) of the ferric chloride, (B) of the stannous 
chloride ? 

(4) Optional. The value of k for the reaction between iodine and hydrogen 
gas at 440 is 0.0005, where velocity is expressed in gram molecules of 
hydrogen used up, per liter, per second. What will be the velocity of this 
reaction when 0.1 gram molecule of hydrogen is mixed with 0.01 gram 
molecule of iodine in a liter volume at 440 ? 



Experiment XIV. 

\ 

Chemical Equilibrium. 

Discussion. Frequently substances which react readily enough when 
brought together, nevertheless do not react completely; that is, the reaction 
comes to a halt while some of the unchanged original substances still remain. At 
least, this often happens when the products of the reaction are not removed as 
fast as they are formed; if they are removed in this way, every reaction will 
ultimately go to completion. 

The obvious explanation of these incomplete reactions is that the substances 
formed, unless one or more of them is immediately removed from the field of 
action, at once begin to react backward to re-form the original substances. The 
velocity of the direct reaction of course becomes progressively less as the original 
substances are used up. On the other hand, the velocity of the reverse reaction 
becomes progressively greater as the concentrations of the reaction products 

31 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

increase. Ultimately these two reactions, the direct and the reverse, acquire equal 
velocities, and a stationary condition of equilibrium is attained, where the origi- 
nal substances and the reaction products are both present. These incomplete 
reactions then represent chemical equilibria. 

In such equilibria, if we attempt to alter the condition of equilibrium by 
adding one or more of the original substances to the equilibrium mixture, the 
direct reaction will proceed with greater rapidity, and more of the reaction pro- 
ducts will accumulate until finally the reverse reaction becomes as rapid as the 
direct reaction, and equilibrium is again attained. In other words, the addition 
of any of the original substances will cause the equilibrium to shift in a direction 
to remove the substance added, and so to oppose the attempt we have made to 
alter the condition of equilibrium. Similarly, addition of the reaction products 
will cause the equilibrium to shift in a direction to remove these added substances 
and so again to oppose our attempt to alter the condition of equilibrium. 

A chemical equilibrium, therefore, behaves just like a mechanical one: for, if 
we attempt, for instance, to pull downward a weight hanging on a coiled spring, 
the weight will move in such a direction as to increase the pull of the spring and 
so to oppose our attempt to alter the condition of equilibrium. 

A useful algebraic statement of the above conclusions is easily obtained from 
the principle which was found to govern the effect of concentration on the velocity 
of chemical reaction. Applying this principle to the incomplete reaction 

a + b + c ^ ai + bi + Ci . 



the velocity, V D of the direct reaction, that is, of the reaction toward the right, 
would always be 

V D = k D X C a X C b X C c 

where C a , C b , C c , etc., are the concentrations of the substances on the left of the 
equation. The velocity of the reverse reaction would similarly be 

V R = k R X C ai X C bl X C ei ...... 



where C ftl , C bl , C Cl , etc. are the concentrations of the substances on the right of 
the equation. 

When the stationary condition of equilibrium is attained, the direct and the 
reverse reaction will be taking place with equal velocity; that is, at equilibrium , 

V D = V R 
and therefore 

k D X C a X C b X C c .... = k R X C ai X C bl X C C] .... 
or dividing through 

V-^aj X Cvbj X V^ C j 



C a X C b X C c 

32 



= K. 



In words, this expression tells us that the product of the concentrations of the 
substances formed in a chemical reaction, divided by the product of the concen- 
trations of the substances consumed, is at equilibrium, equal to a constant. It is 
called the Concentration Law and furnishes the simplest means of computing the 
effect of any change in concentration on the condition of an equilibrium. 

The reaction between ferric chloride and ammonium thiocyanate, studied in 
the following experiment, is an instance of an incomplete reaction or a chemical 
equilibrium. It is a favorable one for study, because any change in the concen- 
tration of the ferric thiocyanate, which is one of the products of the reaction, can 
easily be recognized by the change in the red color of the solution. 

Directions. 

(a) Place 10 c. cm. of water in each of two test-tubes, add a single drop of 
ferric chloride solution to one, and a single drop of ammonium thiocyanate 
solution to the other. Now mix the solutions (Ei). What evidence is 
there that interaction has taken place ? 

(6) Divide the mixture from (a) equally into four test-tubes. Keep one for 
comparison; to the second add one drop of ferric chloride solution; to the 
third, one drop of ammonium thiocyanate; and to the fourth add a few 
drops of ammonium chloride. What do these results prove regarding the 
original reaction (Ei) ? Explain these results on the basis of reaction 
velocity. 

Outside Questions. 

(1) Write the algebraic expressions for the concentration law as applied to the 
above equilibrium. Explain the effects of the addition of ferric chloride, 
ammonium thiocyanate, and ammonium chloride. 

(2) What would be the effect of adding an amount of solid ferric thiocyanate 

to the equilibrium mixture just equal to that already present ? 



Experiment XV. 
The Electrical Conductivity of Solutions. 

Discussion. The abnormally great depressions of the freezing point shown 
by aqueous solutions of strong acids, strong bases, and salts, indicate that these 
substances break up in solution into smaller molecules or separate atoms. The 
following experiments are designed to bring out another peculiarity of these same 
solutions which points to the same conclusion, and in addition shows that these 
smaller molecules and separate atoms are electrically charged. 

33 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Directions. 

(a) Obtain, on temporary order from the store-room, a conductivity appara- 
tus 1 , including electrodes, electrode-holder, socket, lamp-cord, and plug, and 
attach it to the screws on your bottle rack. Unscrew the lamp (a 20 to 40 
watt tungsten) suspended over your desk, insert the plug in its place, and 




Fig. 3. 

screw the lamp into the socket of your apparatus. Fit the electrodes in 
their place and dip the copper wires into the mercury cups. Examine the 
wiring of your apparatus, draw a diagram of it, and observe that the lamp 
is connected in series with the electrodes. Convince yourself that such is 
the case by short-circuiting the two electrodes with a knife-blade or some 

1 (See Fig. 3.) The electrodes in this apparatus consist merely of two fine platinum wires, 
each sealed into a small, thick-walled glass tube. The glass tubes are fitted at the top into a two- 
hole rubber stopper. Their lower ends are allowed to collapse in a blast flame to form solid rods, and 
these rods are sealed together at their ends. The platinum wires are wound spirally around the rods, 
and their free ends are attached to the rods by a little low-melting sealing glass. This apparatus is 
simple and cheap, and when properly made, is robust. 

34 



EXPERIMENT XV 

other metallic object. If the lamp is in series, the brilliancy of its light will 
be a measure of the electrical conductivity of the short-circuiting sub- 
stance. If the electrodes are not clean and dry, rinse them off in a test- 
tube full of water, followed by one full of alcohol. 

Place 20 c. cm. of each of the following liquids in as many medium- 
sized test-tubes, and label each. The aqueous solutions are to be prepared 
by further dilution, if necessary, of the dilute solutions on the shelves. 

(1) Toluene. Use a perfectly dry test-tube for this liquid. 

(2) A solution of hydrogen chloride in toluene. Use a perfectly dry test- 
tube for this liquid also, and keep it stoppered until it is required. 

(3) Glacial (99.5 %) acetic acid. Use a dry test-tube. 

(4) Water. 

(5) 0.1 M sugar solution. 

(6) 0.1 M hydrochloric acid solution. 

(7) 0.1 M acetic acid solution. 

(8) 0.1 M sodium hydroxide solution. 

(9) 0.1 M ammonium hydroxide solution. 
(10) 0.1 M ammonium acetate solution. 

Measure the conductivity of each of the above liquids by submerging the 
electrodes in them, in the order given. After the toluene solution (No. 2) 
has been measured, any liquid adhering to the electrodes must be rinsed 
off with a test-tube full of alcohol. After the other liquids have been 
measured, rinse them off with a test-tube full of water. Record your 
results in tabular form. Save the liquids for (6) and (c). When finished, 
return your conductivity apparatus at once to the store-room. 

(6) Compare the action on litmus paper of the above 0.1 M solutions of 
hydrochloric and acetic acids. Place one drop of each liquid on the 
tongue, and compare the sour tastes. Similarly, compare the actions on 
litmus and on the tongue of the 0.1 M solutions of sodium and ammonium 
hydroxides. 

(c) In a dry test-tube try the action of some of the toluene solution of 
hydrogen chloride on a piece of magnesium ribbon. Then add a single 
drop of water to the mixture. 

Outside Questions. 

(1) From your experiments with toluene and the solution of hydrogen chloride 
in toluene, what do you conclude (A) as to the electrolytic dissociation of 
hydrogen chloride in toluene? (B) as to the nature of the action of an acid 
on a metal? Write an equation expressing this latter conclusion. 

(2) What do your experiments indicate regarding (A) the relative degrees 
of dissociation of hydrochloric and acetic acids ? (B) of sodium and am- 
monium hydroxide solutions ? (C) of salts as compared with weak and 
with strong acids ? 

35 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(3) Compare your results on the conductivities of solutions with the percent- 
age dissociations in these solutions as given in the following table, calcu- 
lated from the depressions of their freezing points. 

PERCENTAGE DISSOCIATIONS OP 0.1 M SOLUTIONS AT 18. 

HC1 92. NaOH 91. 

CH 3 COOH 1.3 NH 4 OH 1.3 

CH 3 COONH 4 72. C 12 H 22 O U 0. 

What regularity do you find ? 

(4) Name the chief four substances present in any aqueous solution of hydro- 
chloric acid. Compute on the basis of the above table the number of grams 
of each present in 10 c. cm. of the 0.1 M solution. Assume the specific 
gravity of this solution to be equal to that of water. 



36 



Experiment XIV. 

Chemical Equilibrium. 
i 

LABORATORY NOTES. 

Name Section ... 

Date ... Desk No. 



Experiment XV. 
The Electrical Conductivity of Solutions. 

LABORATORY NOTES. 

Name , Section ... 

Date ... Desk No. 






Experiment XVI. 
Ionic Equilibria. 

Discussion. Chemical equilibria involving ions usually adjust themselves 
with extreme rapidity, for the speed of ionic reactions as a rule is very great; 
indeed ionic reactions often appear to be instantaneous. Ionic equilibria also are 
very important, for a great many chemical reactions occur in aqueous solutions, 
and this usually means that ions are involved. The simpler ionic equilibria can 
be classified as follows : - 

(a) Dissociation of a single electrolyte. This represents the simplest type of 
ionic equilibrium. Even this type, however, is bound to be affected by 
changes in concentration, for a single molecule of any electrolyte will dis- 
sociate into at least two ions, and the speed with which these recombine 
will evidently vary with the concentration more than will the speed of dis- 
sociation. In other words, dilution of such a solution must increase the 
degree of dissociation ; while evaporation, that is, increase in concentra- 
tion, must decrease the degree of dissociation. 

(6) Equilibria involving two electrolytes giving a common ion. The next type 
of ionic equilibrium in order of simplicity is when two electrolytes, each 
giving rise to a common ion (NaCl and KC1, for instance) are dissolved in 
the same solution. The actual concentration of the common ion is evi- 
dently the sum of the two separate concentrations produced by each 
electrolyte. It follows that the speed of recombination of the ions of each 
electrolyte will be increased by the presence of the other. Any electrolyte 
is, then, less dissociated in the presence of an electrolyte giving a common 
ion than would otherwise be the case. 

(c) Equilibria involving two electrolytes giving a common ion when the solution 
is saturated with respect to one of the electrolytes. If, to a saturated solution 
of one electrolyte, another electrolyte giving a common ion is added, the 
first effect will be to lessen the degree of dissociation of the first electrolyte, 
as shown in (6). This means that additional neutral molecules will be 
formed, and these molecules, as the solution is already saturated with re- 
spect to them, will precipitate. The total solubility of an electrolyte is 
therefore decreased by the addition of an electrolyte giving rise to a com- 
mon ion. 

(d) Equilibria involving two electrolytes giving no common ion. When electro- 
lytes giving no common ion are simultaneously dissolved in the same 
solution, the resulting system is more complicated. Indeed, with only two 

37 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

binary electrolytes, such as KC1 and NaNOs for instance, four separate 
equilibria are involved. The following equations represent these equilibria 
graphically: - 

K + +cr 



tl ti 
KNO 3 NaCl. 

Not only do both of the original salts dissociate into their separate ions, but 
some of the potassium and nitrate ions combine to form potassium nitrate, 
and some of the chloride and sodium ions combine to form sodium chloride. 
There will be then, all told, eight different kinds of molecules and ions 
present, not counting the molecules of water and the ions into which it 
dissociates. 

In a mixed solution of potassium chloride and sodium nitrate such as this, 
if the two salts are taken in equivalent amounts, the neutral molecules will 
all have about the same concentration, and the concentrations of the ions 
will also be nearly identical among themselves, for all of these salts are 
easily soluble and are about equally dissociated. If, however, any of the 
neutral molecules are volatile, or sparingly soluble, or but slightly disso- 
ciated, then these substances will tend to form at the expense of the others 
and the even balance will no -longer be maintained. How completely this 
balance shifts will depend on how volatile, how sparingly soluble, or how 
slightly dissociated the product may be. 

Directions. 

(a) Add a little potassium bromide to some water in a test-tube. What is the 
color of the bromide ion ? Take a small quantity, say 0.2 gram, of cupric 
bromide in a small, dry test-tube. Add two drops of water and agitate for 
some time. Then add more water, a drop or two at a time, shaking vig- 
orously and giving the substance time to dissolve if it will, after each 
addition. Continue the addition of water until the substance has all 
dissolved, and afterward until the change in color is complete. What is 
the color of the cupric bromide molecule ? What is the color of the cupric 
ion ? Compare the latter color with that of a dilute cupric sulphate 
solution (?). Explain the changes you observed. 

(6) Repeat (a) using a fresh portion of cupric bromide, but stop the addition 
of water at the green stage. Divide the mixture into two parts; to one 
add 2-3 grams of potassium bromide and shake vigorously (?); to the 
other add 4-5 grams of cupric chloride and shake vigorously (?). Dilute 
both portions with water. Interpret your results. 

(c) Place 2 c. cm. of a saturated solution of silver acetate in each of two test- 
tubes. Add to one tube a small crystal of silver nitrate (not more than 

38 



Experiment XVI. 
Ionic Equilibria. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



EXPERIMENT XVI 

/ 

0.05 gram). Agitate the solution until the crystal has completely dis- 
solved. If no change is noted at once, set the tube aside for a few minutes 
and observe again. To the second portion add a small crystal of sodium 
acetate and observe as before. Interpret your results. 

(d) Heat a few cubic centimeters of sodium acetate solution and observe 
whether it has any odor. Add a few drops of any strong acid, and again 
observe whether there is an odor. If you are in doubt as to the cause of the 
odor, smell some glacial acetic acid. Knowing that acetic acid is a weak 
acid, interpret your results. 

(e) Prepare 20 c. cm. of a 0.1 M hydrochloric acid solution, and divide 
it equally into two test-tubes. To one test-tube add 3 c. cm. of M 
sodium acetate solution. Compare the sour tastes of the two solutions, 
using a glass rod, placing first a drop of the one solution and then a drop 
of the other on the tongue. This, according to Experiment XV, was 
an approximate method of comparing the hydrogen ion concentrations 
of different solutions. Explain your results from the point of view of 
equilibria. 

Outside Questions. 

(1) Write equilibria equations showing the equilibria involved in each of the 
above experiments. 

(2) Write equilibria equations in the same way for the interaction of solutions 
(A) of silver nitrate and sodium chloride, and (B) of sodium hydroxide 
and hydrochloric acid. 



Experiment XVII. 

The Titration of Acids and Bases. 

Discussion. The concentrations of two solutions often are compared by 
measuring the volume of one solution required to react completely with a given 
volume of the other. This process is called titration. 

That this method can be applied to the comparison of acids and bases really 
depends on the nature of the electrolytic dissociation of water. We know that 
this dissociation is very slight indeed, for pure water is an extremely poor con- 
ductor of electricity. What little dissociation does occur is represented by the 
equation 

H 2 H + + OH~. 

If an electrolyte dissociates slightly, its ions evidently tend to recombine 
very completely, and hence only small amounts of each can remain uncombined 
in the same solution. When the component ions of such an electrolyte are added 

39 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

to a solution, very complete combination will occur. When, therefore, a sub- 
stance furnishing hydrogen ions (an acid) is added to a substance furnishing 
hydroxyl ions (a base) in aqueous solution, the hydrogen and hydroxyl ions com- 
bine very completely to form water. 1 The other ions derived from the acid and 
base combine only partially, for the salt they thus form dissociates freely. 

When acid and base are mixed in exactly equivalent amounts, both hydrogen 
ions and hydroxyl ions will practically disappear. The solution is then neither 
acid nor basic, but neutral. It is fortunately very easy indeed to tell when this 
neutral condition is attained, for there are many substances which may be 
added whose colors are decidedly different in acid and in basic solutions. If a 
little of one of these " indicators " is dissolved, say in a solution of a base, and 
an acid solution is then added slowly, a point will finally be reached when a 
single additional drop of acid will produce a marked change in color. This change 
indicates the point where the last of the excess of hydroxyl ions has been removed 
and an excess of hydrogen ions has begun to accumulate. 

At this neutral point, exactly equivalent amounts of acid and base must have 
reacted, so if the volumes of both solutions which have been mixed are known, it 
is easy to compute the concentration of one with respect to the other. They are, 
of course, simply inversely proportional to the volumes. 

The convenience of this method of comparison depends largely on the great 
rapidity with which these equilibria readjust themselves, that is, on the great 
velocity with which water dissociates and on the very much greater velocity with 
which hydrogen and hydroxyl ions recombine. 

1 Just how completely hydrogen and hydroxyl ions combine, or in other words, exactly what concen- 
trations of hydrogen and hydroxyl ions can simultaneously be present in an aqueous solution, may be 
readily calculated from the following considerations : 

The concentration law, applied to the dissociation of water, is 

C/H X C OH ~ _ _, 

= J\ H ,0- 



-'H.jO 



The value of K^o, the equilibrium constant for this dissociation, has been found in a number of ways to 
be 0.000,000,000,000,000,18 or 1.8 X lO" 16 . The molar concentration of water in pure water, or in any 
dilute aqueous solution, is of course the number of 'molecular weights of water expressed in grams, that 
is, gram molecules, in a liter of the solution. It is then ijga or 55.6, approximately. Substituting these 
values in the above equation 

'-'n X V-'OH 

- L8 x 10 " 16 



we have C H + X Con' = 55.6 X 1.8 X 10- 16 = 1 X IQ" 14 . 

That is, in pure water or any aqueous solution, the product of the concentrations of the hydrogen and 
hydroxyl ions, or the ion product of water, as it is called, is equal to 1 X 10~ 14 . In pure water, where 
both ions must be present in equal amounts, the concentration of each must be 1 X 10~ 7 ; in 0.01 M 
hydrochloric acid, assuming complete dissociation, the concentration of the hydrogen ion would be 
1 X 10- 2 , that of the hydroxyl ion 1 X 10~ 12 . 



40 



EXPERIMENT XVII 

Directions. 

(a) Prepare 250 c. cm. of an approximately molar solution of sodium hydrox- 
ide, weighing out the sodium hydroxide in a beaker on the rough balances 
of the laboratory. Secure an equal volume of the standard solution of 
hydrochloric acid from the end-desk shelves. Secure a pair of burets on 
temporary order from the store-room, and fill one with the sodium hydrox- 
ide solution and the other with the hydrochloric acid solution, first rinsing 
out each buret twice with a little of its solution. Place 100 c. cm. of water 
in a clean, medium-sized beaker, add a few drops of phenolphthalein solu- 
tion to serve as an indicator, read the upper levels of both liquids carefully 
to 0.01 c. cm., record your readings, and allow about 40 c. cm. of acid to 
run into the beaker. Now allow the alkali (the base) to run in, at first 
rapidly, and then, as the neutral point is approached, very slowly, until 
the faintest permanent color appears. Convince yourself that the exact 
end-point has been reached by making sure that a single drop of the acid 
destroys the color. Restore the faint permanent color, and again read the 
levels. Refill the burets and make a second titration. If this does not 
agree with the first titration within the limits of unavoidable error, repeat 
again. From the average of your concordant results calculate the con- 
centration of your sodium hydroxide solution. 

(6) Problem. Having established the concentration of your sodium hydrox- 
ide solution, apply to your instructor for a sample of acid of unknown 
concentration and determine its concentration by titration against your 
sodium hydroxide solution. Report your result to the instructor. 

Outside Questions. 

(1) Sodium hydroxide is about 73% dissociated at molar concentration. 
Calculate the concentration of the Na + , OH~, NaOH and H + in this 
solution. 

(2) How many cubic centimeters of 1 M hydrochloric acid would be required 
to neutralize 50 c. cm. of 0.01 M calcium hydroxide solution ? 

Experiment XVIII. 
Hydrogen Sulphide. 

Discussion (R 371-372). Sulphur resembles oxygen more closely than does 
any other element. In studying hydrogen sulphide, therefore, one should look 
for comparisons between it and hydrogen oxide, or water. 

Directions. 

(a) Properties of the Gas. 

Connect a glass nozzle with the laboratory supply of hydrogen sulphide 
and set fire to the jet of gas (Ei). Note the color of the flame, and com- 

41 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

pare its odor with that of the unburnt gas, and with that of burning 
sulphur. Hold a porcelain dish in the middle of the flame for a few 
moments (?). What substances must be present in an uncombined state 
in the interior of the flame? What do these facts indicate regarding the 
stability of hydrogen sulphide when heated ? 

(6) Properties of the Aqueous Solution. 

(1) Saturate 15 c. cm. of water in a test-tube with hydrogen sulphide gas. 
Test the solution with litmus paper. Dilute a little potassium perman- 
ganate solution in a test-tube, add dilute sulphuric acid, heat to boiling, 
and add the solution of hydrogen sulphide till no further action occurs. 
Manganous sulphate (MnS0 4 ) is formed, a soluble and nearly colorless 
salt (E 2 ). What is the precipitate ? 

(2) Add a few drops of an iodine solution to a little of the hydrogen sul- 
phide solution (E 3 ). What does this show regarding the relative affinities 
of sulphur ions and of halogen ions for negative electricity ? 

(3) Allow the remainder of the hydrogen sulphide solution to stand in con- 
tact with the air until the next period (E 4 ). Also allow a little hydrogen 
iodide solution to stand exposed in the same way (E 5 ). What do (2) 
and (3) show regarding the relative affinities of oxygen, sulphur, and iodine 
for negative electricity ? 

(4) Optional. In five test-tubes dilute about 1 c. cm. of each of the 
following solutions with 10 c. cm. of water and pass in hydrogen sulphide 
gas: (1) copper sulphate; (2) cadmium sulphate; (3) zinc sulphate; (4) 
nickel sulphate; (5) sodium sulphate. If no precipitate forms, first add 
a little ammonium hydroxide solution and then pass in more hydrogen sul- 
phide. Explain the different behaviors of these salts (R 375). 



42 



Experiment XVII. 
The Titration of Acids and Bases. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XVIII. 
Hydrogen Sulphide. 

LABORATORY NOTES. 

Name Section .... 

Date Desk No. 



Experiment XIX. 

Hydrolysis. 

Discussion. In the previous discussion of ionic equilibria (Experiment 
XVI), any action of the ions of water itself on the ions of the dissolved electro- 
lyte was neglected. This is usually legitimate, for water dissociates so very 
slightly into its ions that under ordinary conditions their concentrations are 
almost vanishingly small (lxlQ- 7 M in pure water). Sometimes, however, even 
these concentrations produce a noticeable effect on the ionic equilibria. 

When sodium acetate, for instance, is dissolved in water, it dissociates 
extensively, like other salts, forming sodium and acetate ions; 

CHsCOONa Na+ + CH 3 COO~ 

Acetic acid, however, is a weak acid, which means, of course, that it dissociates 
but slightly, and therefore that only small concentrations of acetate ions and 
hydrogen ions can remain free, uncombined, in the presence of each other. 
There are, to be sure, but very few hydrogen ions present in the water, but the 
acetic acid is so weak that some of these combine with the acetate ions to form 
undissociated acetic acid. The equilibria thus established may be represented 
by the following scheme : - 

CH 3 COONa 




Some of the sodium and hydroxyl ions must also combine, as indicated, but 
this will occur only to a very limited extent, for sodium hydroxide is a strong 
base and therefore highly dissociated. 

It follows from these considerations that there will be more hydroxyl ions 
than hydrogen ions in solution, for at the start both were present in equal 
amounts. The solution will, therefore, react alkaline. 

Directions. 

(a) Dissolve a crystal of sodium sulphide in water and test the solution with 
litmus paper. Write equilibria equations showing how the observed 
effect was produced. 

(b) Test a solution of ferric chloride in the same way. Formulate equilibria 
equations as before. 

43 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(c) Test a solution of ammonium chloride in the same way. What do you 
conclude regarding the weakness of acid or base required to produce a 
pronounced hydrolysis ? Moisten a strip of litmus paper with ammo- 
nium chloride solution, and heat it for a few moments in a current of 
steam from the steam bath. What do you conclude is the effect of in- 
creased temperature on hydrolysis ? 

Outside Questions. 

(1) What relation does hydrolysis bear to neutralization ? 

(2) What should you predict regarding the hydrolysis of a solution of 
(A) potassium borate ? (B) barium perchlorate (R 330) ? 

(3) What will happen when a salt of a weak acid and a weak base is dissolved 
in water ? How would such a solution behave toward indicators ? 

(4) Optional. Can you see any connection between the effect of increased 
temperature on hydrolysis and a phenomenon accompanying neutraliza- 
tion (R 260) ? 



Experiment XIX. 
Hydrolysis. 

LABORATORY NOTES. 

Name Section ... 

Date ... Desk No. 



Experiment XX. 

Sulphurous Acid. 
Discussion (R 378). 

Directions. 

(a) Pass a current of sulphur dioxide from a cylinder of the liquified gas into 

about 15 c. cm. of water for a half a minute (Ei). Divide the solution 

into four portions. 
(6) Immerse a strip of blue litmus paper into one portion and add dilute 

ammonium hydroxide to faint alkalinity (E 2 ). Then add a few drops 

of barium chloride solution (E 3 ), followed by a few drops of hydrochloric 

acid (E 4 ). 

(c) To a second portion, add bromine water (E B ) until the color is permanent; 
then continue as in (6). 

(d) To a third portion add an equal volume of dilute sulphuric acid and then 
add a solution of potassium permanganate until no further action* occurs 
(E 6 ). What property of sulphurous acid do (c) and (d) illustrate ? 

(e) To a fourth portion add a few drops of stannous chloride solution and 
heat (?) (E 7 ). Smell the hot solution. Hold a strip of filter paper, moist- 
ened with a solution of lead salt, in the mouth of the test-tube, and con- 
tinue heating (E 8 ). What property of sulphur dioxide does this illustrate ? 

(/) Dissolve a crystal of sodium sulphite in water, and test the solution with 
litmus paper. What does this demonstrate regarding the strength of 
sulphurous acid ? 

Outside Questions. 

(1) Write a series of connected equilibria equations showing what occurs 
when a solution of ammonium sulphite is treated with a solution of 
barium chloride, followed by one of hydrochloric acid. 

Experiment XXI. 

Sulphuric Acid. 
Discussion (R 382-393). 
Directions. 

(a) Dip a splint into dilute sulphuric acid and make marks on a piece of 
paper. Dry both splint and paper over a low flame or on a radiator (?) 
(Ex) (R 388). 

(6) Pass a current of hydrogen sulphide through 2-3 c. cm. of concentrated 
sulphuric acid. Note the odor and the precipitate (E 2 ). 

45 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(c) Heat a little charcoal with concentrated sulphuric acid (?) (E 8 ). What 
property of sulphuric acid is illustrated by both (6) and (c). Why does 
dilute sulphuric acid exhibit this property to but a slight degree ? 

(d) Add a drop of sulphuric acid each to separate solutions of a lead salt, a 
barium salt, and a calcium salt, of approximately tenth molar concen- 
tration. 

Outside Questions. 
(1) Enumerate the common insoluble sulphates (R 544). 



Experiment XX. 
Sulphurous Acid. 

LABORATORY NOTES. 

Name Section .... 

Date Desk No. 



Experiment XXII. 
Catalysis. 

Suggested Reading. R. K. Duncan, The Chemistry of Commerce, pp. 
15-40. 

Discussion. It was shown in Experiment XIII that temperature and 
concentration are important factors in determining the velocities of chemical 
reactions. Chemists have also found that small amounts of apparently unessen- 
tial and indifferent impurities often greatly hasten chemical reactions. Thus, 
small quantities of manganese dioxide greatly accelerate the evolution of oxygen 
gas from molten potassium chlorate; finely divided platinum promotes in a 
marked degree the combination of sulphur dioxide with oxygen gas; ferrous 
salts act in a similar fashion upon the reaction between solutions of hydrogen 
iodide and hydrogen peroxide; etc., etc. The manganese dioxide, the finely 
divided platinum, and the ferrous salts are not consumed in these reactions, and 
therefore a small quantity is able to hasten the chemical reaction of an indefinite 
amount of material. Substances which act in this way are called catalytic agents, 
or catalysts; the phenomenon itself is called catalysis. 

Catalysis, in many cases, appears to be as much a physical as a chemical phe- 
nomenon. For instance, instead of manganese dioxide almost any finely divided, 
infusible material may be employed to hasten the evolution of oxygen from po- 
tassium chlorate. The more finely divided the infusible material is, the more pro- 
nounced is its accelerating effect. Apparently, the decomposition of the molten 
chlorate occurs more rapidly on the surface, and so the greater the surface, the 
more rapid the reaction. The action of the catalyst thus seems to be merely a 
physical one. 

In most cases, however, we are certain that catalysis is a chemical, rather 
than a physical phenomenon, that is, that the catalyst does take part in the 
chemical reaction, although it is not permanently destroyed in the process, being 
re-formed as fast as it is used up. In the lead-chamber process, for instance, the 
oxides of nitrogen react with sulphur dioxide, water, and oxygen to form nitro- 
sylsulphuric acid, and this, with more water vapor, forms sulphuric acid, and 
along with it, the original oxides of nitrogen. There are, to be sure, two steps in 
this indirect formation of sulphuric acid, but each of them is much more rapid 
than the direct combination of sulphur dioxide with water and oxygen to form 
sulphuric acid, and therefore the reaction as a whole is much more rapid. A great 
number of catalytic effects are known to occur in a similar way. 

Sometimes, the addition of an indifferent, foreign substance retards rather 
than hastens a reaction. The added substance behaves, in other words, as a nega- 

47 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

live catalyst. In every case which has been carefully studied, it has been found 
that the negative catalyst really does not directly affect the reaction at all. It 
merely destroys some ordinary (positive) catalyst already present. 

Finally, many cases are known where reactions are greatly hastened by other 
reactions simultaneously taking place in the same solution. For instance, in di- 
lute solutions, ammonia is oxidized very slowly indeed by potassium permanga- 
nate; sodium hypochlorite on the other hand, under these conditions, oxidizes 
ammonia rapidly. If, now, sodium hypochlorite is added to a dilute solution of 
ammonia and potassium permanganate, not only does the sodium hypochlorite 
oxidize the ammonia, but the potassium permanganate does likewise. Similarly, 
indigo is oxidized very slowly to indigo-white, and ferrous salts are oxidized 
rapidly to ferric, by hydrogen peroxide. If ferrous salts are added to a solution 
of indigo and hydrogen peroxide, not only are the ferrous salts rapidly oxidized, 
but so also is the indigo. One reaction evidently induces or stimulates the other,' 
and such reactions are called either "induced," or " sympathetic" reactions. 

The explanation of this peculiar phenomenon is that the naturally rapid re- 
action produces a substance which attacks the otherwise slowly reacting sub- 
stance. In the case of the ammonia, potassium permanganate, and sodium hypo- 
chlorite, for instance, the ammonia and hypochlorite produce a substance which 
reacts rapidly with the potassium permanganate. 

Directions. 

(a) To show the small amount of catalyst required. Mix 25 c. cm. of a 0.2 M 
ammonium oxalate solution with 10 c. cm. of mercuric chloride solution of 
the same concentration, and divide into two portions. Prepare also 50 
c. cm. of a 0.001 M solution of potassium permanganate. Add two drops of 
the latter solution to each of the above portions, and warm both portions 
simultaneously in the steam bath, observing in which portion the precipi- 
tate of mercurous chloride first makes its appearance (Ei). 

(6) Negative Catalysis. Pass sulphur dioxide gas into 100 c. cm. of distilled 
water until one cubic centimeter of the solution diluted with water requires 
approximately one cubic centimeter of 1 M sodium hydroxide solution for 
neutralization, using phenolphthalein as indicator. Add 1 c. cm. of this 
solution to 25 c. cm. of water in each of three large test-tubes. To one 
of these solutions add one drop of copper nitrate solution; to a second, 
add a few drops of alcohol. Then add to each test-tube, as rapidly as 
possible, 2 c. cm. of 1 M sodium hydroxide solution, close each with a 
cork stopper, and shake at intervals for five minutes. Then to each solu- 
tion add a few drops of concentrated hydrochloric acid, followed by a 
little barium chloride solution. Explain the different behaviours of these 
solutions. 



48 



EXPERIMENT XXII 

(c) Induced Reactions. Mix 5 c. cm. of nickel chloride solution with 10 c. cm. 
of dilute sodium hydroxide solution in 150 c. cm. of water in a 500 c. cm. 
Erlenmeyer flask. To a little of the mixture in a test-tube add bleaching 
powder, and shake (E 2 ) (R 761). Close the flask, and shake for a few 
moments. Add a few drops of dilute sulphurous acid, and again shake 
for a few moments (E 3 ). Then allow the mixture to stand undisturbed 
for a few minutes (E 4 ). This process may be repeated indefinitely. 

Outside Questions. 

(1) Compute the weight of potassium permanganate present in the two drops 
(about 1/5 c. cm.) of solution which was added in (a). 

(2) Suggest a possible explanation for the behavior observed in (c) . 



Experiment XXIII. 
Properties of Free Bromine and Iodine. 

Discussion. The elements of the halogen group show a very marked family 
likeness. This can be seen best by tabulating the physical and chemical proper- 
ties of these elements (R 226). The following experiments will serve further to 
emphasize this fact in so far as bromine and iodine are concerned. 

Directions. (Caution! Be extremely careful not to get bromine on the hands 

or face. If you do, wash it off instantly with water.) 

(a) Pour a few drops of bromine (hood!) into a dry test-tube and warm 
very gently (?). Cool, add 5 c. cm. of water and shake. Smell this 
mixture guardedly. Does it smell like chlorine ? Finally, add a pinch 
of sodium bromide and shake again. Dilute a drop of this solution 
in another test-tube with 10 c. cm. of water, add a few drops of carbon 
disulphide, and shake. 

(6) Repeat (a) using a few crystals of iodine in place of the bromine, and 
sodium iodide in place of the bromide. Afterwards prepare a solution of 
starch by grinding up a gram of starch with one or two cubic centimeters 
of water in a mortar, and pouring this emulsion into a beaker full of 
boiling water. Add a little of this starch solution to another beaker 
full of water, stir and add a drop or two of the above solution containing 
iodine and potassium iodide. 

Outside Questions. 

(1) What is the explanation of the influences of potassium bromide and 
potassium iodide on the solubilities of bromine and iodine, respectively, 
in water (R 235) ? 

49 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(2) What do you conclude from the above experiments, regarding the relative 
solubilities of iodine in water and in carbon disulphide? State the law 
of partition (R 155-156) as applied to this case. 

(3) Optional. The partition ratio of iodine between water and carbon disul- 
phide is 1: 585 at 25. If 0.1 gram of iodine is dissolved in a mixture of 
10 c. cm. of water and 1 c. cm. of carbon disulphide, how much iodine 
will be contained in the water ? 



60 



Experiment XXII. 
Catalysis. 

LABORATORY NOTES. 

Name Section ... 

Date ., Desk No. 



Experiment XXIII. 
Properties of Free Bromine and Iodine. 

LABORATORY NOTES. 

Name Section ... 

Date ... Desk No. 



Experiment XXIV. 

Preparations and Properties of Hydrogen Fluoride, Bromide, and Iodide. 

Discussion. Since all of the hydrides of the halogens are volatile substances, 
they can be prepared by treatment of non-volatile halides with suitable non- 
volatile acids; for we have seen that an otherwise incomplete reaction will go 
to completion if one or more of its products are removed. 

Directions. 

(a) The Action of Hydrogen Fluoride on Silica (R241). Mix a little powdered 
fluorite (calcium fluoride) with some sand (silica) in a dry, medium- 
sized test-tube. Cover with 5 c. cm. of concentrated sulphuric acid, shake, 
warm gently (Ei), and hold a glass rod or tube with a drop of water hang- 
ing from its lower end, at a height of one centimeter above the acid (?) 
(E 2 ). After a few moments test this drop with litmus paper. 

(6) Optional. Etching Glass with Hydrofluoric Acid. Melt a thin layer of 
paraffine over the outer surface of the upper third of a test-tube. With 
a sharp point engrave your name on this layer, cutting through to the 
glass. With a small swab made from a piece of cloth caught in a slit on 
the end of a wooden stick, apply aqueous hydrofluoric acid to the en- 
graving. (Avoid breathing the fumes of hydrofluoric acid; they are 
poisonous!) Leave the test-tube in the hood for some minutes, then 
rinse off the acid, and melt and wipe off the paraffine. 

(c) Hydrogen Bromide. Place about one gram of sodium bromide in a test- 
tube; cover it with concentrated phosphoric acid (E 3 ). Warm, if neces- 
sary, to start action. Observe the odor (caution!). Blow across the 
end of the tube. Open a bottle of ammonium hydroxide solution and 
hold its mouth near the mouth of the test-tube (E 4 ). Add a cubic centi- 
meter of concentrated sulphuric acid and warm (E 5 ). Add a pinch of 
manganese dioxide (E 6 ). 

(d) ' Hydrogen Iodide. Repeat (c) using sodium iodide instead of sodium bro- 
mide, but omit the addition of manganese dioxide (E 7 -E 9 ). Identify as 
many products as possible in the reaction with sulphuric acid. 



51 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Outside Questions. 

(1) What property of hydrofluoric acid finds expression both in its formula 
in the gaseous state and in its action on silicon tetrafluoride (R 242 
521 and 527) ? 

(2) Explain the phenomenon you observed when you blew across the mouth 
of the test-tube containing hydrogen bromide gas (R 182). 

(3) What properties must an acid possess to be used advantageously in the 
preparation of hydrogen iodide from sodium iodide ? Boric acid is so 
weak that it scarcely turns litmus red. Could it be used (R 527) ? 



52 



Experiment XXIV. 
Preparations and Properties of Hydrogen Fluoride, Bromide, and Iodide. 

LABORATORY NOTES. 

Name Section 

Date Desk No. . 



Experiment XXV. 
Precipitation Reactions of Halogen Ions. 

Discussion. Certain of the halogen compounds of the metals are sparingly 
soluble, and some of them have distinctive colors. Since they are formed when- 
ever the constituent ions are brought together at adequate concentrations they 
may serve as tests for these ions. 

Directions. 

(a) Place about 5 c. cm. of 0.1 M sodium bromide solution in each of four 
test-tubes. Add to the first a few drops of calcium nitrate solution, 
to the second lead nitrate, to the third mercurous nitrate, and to the 
fourth silver nitrate. Tabulate your results. 

(6) Repeat (a) using sodium iodide solution. 

(c) Repeat (a) using sodium fluoride solution. 

(d) Repeat (a) using sodium chloride solution, unless you remember the ap- 
pearance of the precipitates which would be thus formed. 

Outside Questions. 

(1) Tabulate the precipitates formed in this experiment in the order of their 
solubilities (Smith Inside front Cover). Mercurous chloride, bromide, 
and iodide have molar solubilities respectively of 8 X 10~ 7 , 2 X 10~ 8 , 
and 3 X 1Q- 10 . 

Experiment XXVI. 
The Relative Affinities of the Halogens for Negative Electricity. 

Discussion. A strip of metallic zinc immersed in a solution of a copper 
salt goes into solution and simultaneously precipitates metallic copper. With 
copper sulphate, for instance, in dilute solution, 

Zn + Cu++ + SO" = Zn++ + Cu + SOI". 

The sulphate ion (SO! ~) evidently is not concerned at all in this reaction. In- 
deed, the reaction consists merely in the capture of two charges of positive 
electricity from the copper ion by the zinc atom. This shows that zinc has a 
greater affinity than copper for positive electricity. 

Copper atoms similarly are able to deprive silver ions of their charge, 
while silver atoms in turn produce the same effect upon gold ions. The metals 
can then be arranged in the order of their abilities to precipitate one another 

53 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

from solution, or in other words, in the order of their affinities for positive 
electricity. This arrangement is called the Volta Displacement Series, or the 
Electromotive Series of the Metals (R 361-362). 

It is the object of the following experiments to demonstrate that the halogen 
elements can be arranged in a similar series showing their relative affinities for 
negative electricity. 

Directions. 

(a) Add a drop of carbon disulphide and a few drops of chlorine water to a 
dilute solution of a bromide. Shake (Ei). 

(6) Repeat (a) with bromine water and an iodide (E 2 ). What do you conclude 
would be the effect of chlorine water on an iodide ? Prove it. Write 
equations showing the interchanges of negative electricity in these reac- 
tions. Write the three halogens in the order of their affinities for negative 
electricity. 

Outside Questions. 

(1) How would fluorine act upon a solution containing another halogen ion 
(R 241) ? 

(2) Arrange all four halogens in the order of their affinities for positive elec- 
tricity. How would this series fit into the similarly arranged Volta Series 
of the metals ? What elements would then differ most in this respect 
from fluorine ? 



Experiment XXV. 
Precipitation Reactions of Halogen Ions. 

LABORATOKY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XXVI. 
The Relative Affinities of the Halogens for Negative Electricity. 

LABORATORY NOTES. 

Name Section 

Date ... Desk No. . 



Experiment XXVII. 
Absorption of Iodides by the Stomach. 

Discussion. Most substances which get into the blood from the alimentary 
canal are absorbed through the walls of the small intestine. Only a few sub- 
stances, of which water, alcohol, and the simpler electrolytes are the most im- 
portant, are able to pass at all readily through the walls of the stomach. Of 
these, water and iodides are the only ones rapidly absorbed by the salivary 
glands. Iodides can be easily detected in very small amounts, so that they may 
serve as a means of comparing the rate of absorption by the walls of the stomach 
in different individuals. Unfortunately, the rate of absorption thus measured 
usually has but little significance, for it depends on a number of conditions, such 
as the amount of water, the amount and nature of the food in the stomach, etc., 
which are very difficult to regulate. 

Directions. Test your saliva for iodide by first diluting it with some water 
and then applying the test with starch solution and chlorine water. Swallow 
25 c. cm. of 0.1 M potassum iodide solution, rinse out your mouth and note 
the time. After five minutes, commence testing your saliva for iodide at two 
minute intervals. In this test use somewhat more chlorine water than usual, as 
the saliva appears to destroy the first drops which are added. If absorption is 
normal, a positive test may be expected in fifteen minutes. Compare your rate 
of absorption with that of several more fleshy or several less fleshy members of 
the class. 

Experiment XXVIII. 

The Oxygen Acids of the Halogens. 
Discussion (R 263). 

Directions. 
(a) Hypochlorous Acid and Chloric Acid. 

(1) Soak 10 grams of bleaching powder (" chloride of lime," R 266) in 
50 c. cm. of water for a few minutes with occasional shaking. Meanwhile 
immerse a strip of blue litmus paper in chlorine water and interpret all 
the changes you observe. Filter the extract from the bleaching powder 
through a folded filter. Immerse a strip of red litmus in a 5 c. cm. 
portion of the filtrate; after a few moments acidify with a little dilute 
nitric acid (Ei). Note the smell of the acidified solution. Is it identical 
with the smell of chlorine (E 2 ) (R 268) ? 

55 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(2) To another 5 c. cm. portion of the filtrate in a test-tube, add a drop 
of cobalt chloride solution (E 3 ) (R 760), and while warming very gently 
(E 4 ) insert a glowing splinter into the test-tube. 

(3) Evaporate the remainder (about 35 c. cm.) of the extract from the 
bleaching powder nearly to dry ness on the steam bath. Note the smell 
of the hot, concentrated solution. Transfer the mixture as completely as 
possible to a test-tube, cool, and add dilute nitric acid, a drop at a time, 
with constant shaking until the insoluble carbonate in suspension is just 
dissolved. If you add more acid than is necessary, rinse out the evap- 
orating dish with the solution. This should neutralize the excess of acid. 
Dissolve about a gram of potassium nitrate in the clear solution, which 
should not now exceed 10 c. cm. in volume. Boil for a few moments (E 5 ), 
and then cool persistently under the tap. If no crystals appear, inoculate 
the solution with a single small crystal of potassium chlorate, cool again, 
and shake for a few minutes. 

(6) Hypobromous Acid and Bromic Acid. Add 1 c. cm. of bromine (Caution!) 
to 8 c. cm. of 3 M potassium hydroxide solution (E 6 ). Filter if necessary. 
Immerse a strip of litmus paper in the solution. Transfer a few drops of 
the solution to another test-tube and acidify with dilute sulphuric acid 
(E 7 ). Now boil the remainder of the solution for a few moments, and 
divide into one large and two small portions. Acidify one of the small 
portions as before (E 8 ) . Add a little barium chloride solution to the other 
(E 9 ). Cool the large portion under the tap and save it for (d) (optional). 

(c) Hypoiodous Acid and lodic Acid. Repeat (b) using one gram of iodine in 
place of the t bromine (Ei -Ei 3 ). 

(d) Optional. Purification of Potassium Bromate and lodate. After thor- 
oughly cooling the large portion of the potassium bromate solution ob- 
tained in (6) , transfer the crystals as completely as possible to a small filter 
and drain them free from mother-liquor. Transfer back to a clean test- 
tube, dissolve in the least possible amount of boiling water, again cool thor- 
oughly, and pour off the mother-liquor. Rinse the crystals several times 
with a little alcohol, until some of the waste alcohol, diluted with water, 
gives no indication of bromide ions when treated with silver ions. Scrape 
the crystals on to a filter paper, let them dry in the air, and stopper them 
in a small specimen-tube obtained from the store-room. 

The solution of potassium iodate obtained in (c) may be treated in the 
same way. 



56 



EXPERIMENT XXIX 

Outside Questions. 

(1) What substances are present in chlorine water (R 265) ? 

(2) Write equilibria equations showing (A) what substances are formed when 
bleaching powder is dissolved in water, and (B) why nitric acid is added 
to this solution. 

(3) How do you explain the action of cobalt chloride on the hypochlorite 
solution ? 

(4) Cite two observations from (a) which show that hypochlorous acid is a 
more vigorous oxidizing agent than the hypochlorite ion. 

(5) Compare the actions of heat on solutions of hypochlorites, hypobromites, 
and hypoiodites. 



Experiment XXIX. 
The Relative Oxidizing Powers of the Oxygen Acids of the Halogens. 

Discussion. Some substances are much more vigorous oxidizing agents than 
are others. Nitric acid, for instance, oxidizes many substances upon which 
sulphuric acid has no effect. The oxygen acids of the halogens exhibit similar 
differences, and it is the chief object of this experiment to arrange them accord- 
ing to these differences. 

In such an arrangement, it is helpful to see how the oxidizing powers of 
various oxidizing agents may be connected with the oxidizing power of oxygen 
itself. This connection will appear if we consider the oxidation of some sub- 
stance A by gaseous oxygen. Assume that the oxidation takes place according 
to the equation, 

A + 2 - A0 2 . 

Of course the reverse action is bound to take place to some extent, so that there 
is really an equilibrium established represented by the equilibrium equation, 

A + 2 ^ A0 2 . 
Applying the concentration law to this equilibrium, 

^AOa _ T 



XUo, 

= K X C 0l . 



C A 

But the fraction ^^ evidently represents the fraction of the substance A which 



CAO* 



is oxidized by the gaseous oxygen. In words then this last equation states that 
the fraction of A oxidized is directly proportional to the concentration of the 
oxygen. It follows that we can increase the oxidizing power of oxygen by in- 

57 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

creasing its concentration, or more generally, that by either increasing or de- 
creasing its concentration we can make the oxidizing power of oxygen equal to 
that of any other oxidizing substance. Conversely, we can think of the oxidizing 
power of any substance as simply equal to that of gaseous oxygen at the proper 
concentration. 

We must not think, however, that the velocity with which an oxidation 
occurs always corresponds with the strength of the oxidizing agents. Some 
oxidizing agents, like some human beings, are very strong, but they act slowly. 
The amount of oxidation produced by unit concentration of the oxidizing agent 
when it has done all it can, that is, when equilibrium has been reached, is what 
counts. 

Directions. 

(a) To 5 c. cm. of water in a test-tube add a little carbon disulphide and a 
minute fragment of iodine. Shake vigorously, keeping the mouth of the 
test-tube closed with your thumb. The carbon disulphide should be 
distinctly, but not deeply, colored. Finally add bromine water, a few 
drops at a time, shaking vigorously after each addition (Ei). 

(6) Repeat (a) using a few drops of bromine water in place of the iodine, and 
chlorine water in place of the bromine water (E 2 ). 

(c) To 3 c. cm. of potassium bromate solution in a test-tube add a few drops 
of chloroform and a minute fragment of iodine. Continue the shaking 
for a few moments after all the iodine has dissolved; then add a few 
drops of sulphuric acid and again shake (E 3 ). 

(d) Repeat (c) using a solution of potassium chlorate instead of the potas- 
sium bromate (E 4 ). If no action occurs, warm slightly, or catalyse the 
reaction with a drop of ferrous sulphate solution. 

(e) Repeat (d) using bromine water in place of iodine (E 5 ). 

(/) To about 10 c. cm. of water in a test-tube add a single drop of potassium 
iodide solution and then a single drop of potassium bromide solution. 
Introduce also a few drops of carbon disulphide. Now add chlorine water, 
a little at a time, shaking vigorously after each addition. Continue until 
no further changes occur (E 6 _ 9 ). This procedure is used in analysis for 
recognizing a bromide in the presence of an iodide. 

Outside Questions. 

(1) Deduce the relative oxidizing powers of the above six acids as follows: 
(A} Hypobromous as compared with hypoiodous and hypochlorous, from 
(a) and (6). 

(B) Hypobromous as compared with iodic, and hypochlorous as com- 
pared with bromic, also from (a) and (6). 

(C) Bromic as compared with hypoiodous and iodic, from (c). 

58 



Experiment XXVII. 
Absorption of Iodides by the Stomach. 

LABORATORY NOTES. 

Name Section ... 

Date ... Desk No. 



Experiment XXVIII. 
The Oxygen Acids of the Halogens. 

LABORATORY NOTES. 

Name Section ... 

Date ... Desk No. 



Experiment XXIX. 
The Relative Oxidizing Powers of the Oxygen Acids of the Halogens. 

LABOEATORY NOTES. 

Name Section 

Date ... Desk No. . 



(D) Chloric as compared with the other five acids, from (6), (c), (d), 
and (e). 

Construct a diagram showing these relationships, putting the most pow- 
erful oxidizing agent at the top, with arrows pointing from stronger to 
weaker. What additional data are needed to fix completely the order of 
these acids ? 

(2) How does your arrangement agree with the evidence furnished by thermo- 
chemistry (R 271-272) ? 

Experiment XXX. 
Ammonia and Ammonium Hydroxide. 

Discussion (R 417-421). Ammonia is an unusually interesting substance, 
partly on account of its own properties, but chiefly because with water and 
some other substances as well, it forms the ammonium group, or radical, NH 4 , 
which simulates the atom of a metallic element in a remarkable fashion. 
Ammonium has not been isolated in the free state, but solutions of it in mer- 
cury are readily obtained, and when charged with positive electricity it is 
entirely stable in aqueous solution. Ammonium hydroxide is a relatively weak 
base, and, since ammonia is a gas, is also in effect volatile. This latter prop- 
erty is unusual among inorganic bases. 

Directions. 

(a) Grind 10 grams of ammonium nitrate with 5 grams of soda-lime, or slaked 
lime, in a mortar, and charge the mixture into a test-tube provided with 
a one-hole rubber stopper and a straight delivery tube about 20 cm. long. 
Warm the mixture gently (Ei), guardedly smell the gas evolved, and 
hold a piece of moist red litmus paper in it. Place a dry, inverted test- 
tube over the delivery tube, and warm the generator. After a few moments, 
when the air will largely have been swept out of the inverted test-tube, 
close its mouth with your finger, keeping the test-tube in its inverted 
position, and submerge the mouth in water. Remove your finger, thus 
allowing a little water to enter the tube. Close the tube again and shake, 
then open it again under water. Write equilibria equations showing 
how the ammonia was evolved and absorbed. Save the generator for (6). 

(6) Place one or two small crystals of copper nitrate (R 623) in a narrow, 
dry test-tube, and heat carefully until no further action (E 2 ) occurs, turn- 
ing the test-tube so that the product of the reaction is spread over the 
lower walls. Blow out the gaseous products of the reaction. Lower this 
test-tube over the delivery tube of the ammonia generator, warm the 
generator sufficiently to produce a gentle current of ammonia, and at the 
same time vigorously heat the deposit in the other test-tube (E 3 ) . What 
property of ammonia does this illustrate ? 

59 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(c) Heat a little ammonium chloride in a dry test-tube (?) (E 4 ). Is any 
ammonia given off ? Identify the product on the walls of the tube. 
Test the neutrality of this product. Repeat this experiment using am- 
monium sulphate (E 6 ). Explain the difference in the results. 

(d) Place 50 c. cm. of dilute ammonium sulphate solution in a 250 c. cm. 
flask supported on a retort stand. To this flask fit a safety tube and a 
U-shaped delivery tube opening over 25 c. cm. of water in a small flask. 
Add a few drops of litmus solution to the solutions in each flask. Boil 
the solution in the large flask gently for 5 minutes (?). Explain the 
changes you observe. 

Outside Questions. 

(1) Does the reducing power of ammonia increase or decrease with rising 
temperature (R 419) ? State your reasons. 

(2) What would be the effect of heating (A) ammonium bromide (E 6 ) ; (B) 
ammonium nitrate (E 7 ) (R 450) ; (C) ammonium phosphate (E 8 ) ? 

(3) Why do some substances change into vapor without melting ? What is 
such a change called (R 463) ? 



60 



Experiment XXX. 
Ammonia and Ammonium Hydroxide. 

LABORATORY NOTES. 

Name Section .... 

Date ... Desk No. 



Experiment XXXI. 
Nitric Acid as an Oxidizing Agent. 

Discussion. The extensive use of nitric acid as an oxidizing agent is due to 
a number of causes. In the first place, its action as an oxidizing agent is very 
vigorous. Moreover, the chief reduction products formed from it in such 
action nitrogen dioxide N0 2 , nitric oxide NO, nitrous oxide N 2 0, and am- 
monia NH 3 are gaseous, and therefore easily disposed of. It is also a strong 
acid, and its salts are soluble. Finally, compared with other oxidizing agents 
it is relatively inexpensive. 

Because of its vigor as an oxidizing agent, nitric acid, at least if present 
in excess and if warmed, converts most substances upon which it acts into their 
highest state of oxidation. To what extent it will thereby be itself reduced 
depends on a number of factors, namely, its own concentration, the nature and 
concentration of the reducing agent, the temperature, and also very often on 
the time, for its action is frequently catalysed by various products formed 
during the reaction. 

The effect of most of these factors can be understood without great diffi- 
culty. The effect of the concentration of the nitric acid, for instance, is quite 
what one would expect. The actual concentration of the undissociated nitric 
acid molecule is never large except in very concentrated solutions, for nitric 
acid is a strong acid and highly dissociated in dilute solution. The oxidizing 
power of the nitrate ion is relatively small, for we know that neutral, aqueous 
solutions of nitrates are not strong oxidizing agents. It follows that the oxidizing 
power of nitric acid is due chiefly to its undissociated molecules, and hence the 
oxidizing power must decrease with great rapidity as the acid is diluted. Experi- 
ment shows this to be the case. 

The influence of the nature of the reducing agent is as a rule equally potent. 
The very strong reducing agents, such for instance as the metals near the top 
of the displacement series, will reduce nitric acid very completely, while the 
much weaker reducing agents at the bottom of this series will reduce it only to 
some intermediate stage. 

The following experiments will serve to demonstrate these conclusions. 

Directions. 

(a) Action on a Non-Metal. Add 1 gram of sulphur to 2-3 c. cm. of con- 
centrated nitric acid, and boil for several minutes under the hood. Iden- 
tify as many products of the reaction as you can (Ei). Pour some of the 
clear liquid into another test-tube, dilute with water, and test for the 
sulphate ion. 

61 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(6) Action on Metals. 

(1) Fit a test-tube, 20-25 mm. in diameter, with a safety-tube reaching 
to the bottom of the test-tube, and with a delivery tube opening under 
water in a wide dish. Weigh out two samples of copper turnings of 
about one gram each with an accuracy of 2-3 percent. Charge one of 
these samples into the test-tube, insert the stopper with its two tubes, 
and pour through the safety tube 20 c. cm. of nitric acid prepared by 
diluting 10 c. cm. of concentrated acid with an equal volume of water. 
Warm slightly to start the action (E 2 ), and then collect all of the evolved 
gas over water. Measure its volume and then mix with air (?) (E 3 ). 

Repeat this experiment with the second weighed sample of copper, using 
20 c. cm. of concentrated, instead of the diluted, nitric acid (E 4 ). 'what 
is the gas collected ? Compare the volumes of gas collected in the two 
experiments. If you suspect hydrogen to have been present, investigate 
the effect of diluted hydrochloric acid on copper turnings. What were 
the other products of the reaction, and what became of them ? How do 
you explain the effect of increased concentration of the nitric acid on the 
nature of the reduction products formed ? 

(2) To about 1 gram of zinc in a test-tube, add 10 c. cm. of water, and then 
3 c. cm. of dilute (6 M) nitric acid. Warm, but do not boil the mixture; 
allow it to stand for five minutes (E 5 ). Pour off the solution from the 
undissolved zinc into a small beaker. Make alkaline with a solution of 
sodium hydroxide and warm (?). Suspend a piece of moist red litmus 
paper in the vapor of the warm liquid. How do you explain the difference 
between the action of zinc and of copper on nitric acid ? 

(3) Fit a glass jet into the mouth of a medium-sized test-tube by means of 
a one-hole rubber stopper. Fill the test-tube one-quarter full with dilute 
nitric acid, and dilute it further with twice its volume of water. Add 
some magnesium turnings to the diluted acid, insert the stopper and jet, 
and after the air has been swept from the tube, bring a flame up to the 
opening of the jet (E 6 ). How do you explain the action of magnesium as 
compared with that of zinc on nitric acid ? 



62 



Experiment XXXI. 
Nitric Acid as an Oxidizing Agent. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XXXII. 
Phosphine. 

Discussion. The marked similarity which corresponding compounds of 

nitrogen and phosphorus exhibit is apparent in the case of ammonia and phos- 

These two substances are very similar in physical properties, and there 

are many resemblances in their chemical behavior. Phosphine, however, is less 

stable then ammonia, and is much less soluble in water (R 460-462). 

Directions. [Caution! Perform all experiments with phosphine under the 
Phosphine is somewhat poisonous and is very obnoxious.] Construct a 
generator by fitting a test-tube with a two-hole rubber stopper carrying in one 
hole a right-angle inlet tube extending nearly to the bottom of the test-tube, and 
in the other an inverted U-shaped exit tube with one limb extending just below 
the stopper and the other long enough to reach nearly to the bottom of a second 
test-tube. Place 10-15 c. cm. of water in the generator, and drop into it two 
or three fragments of calcium phosphide not exceeding 3-4 mm. in diameter 
Warm slightly to hasten the action (?) (E,). Add a little hydrochloric acid 
E 2 ). Does phosphine form an acid or a base with water ? Cite observations 
made in this experiment which prove your statements. 

Pass a slow stream of illuminating gas through your phosphine generator, 
while it is still active, and pass the gas through dilute solutions of copper sul- 
phate^(E 3 ), of mercurous nitrate (E 4 ), and of mercuric chloride (E 6 ) (R 654-655). 
Does illuminating gas alone produce these results ? At the completion of the 
experiment add copper sulphate solution to the generator, and wash the con- 
tents down the drain with much water. 

Outside Questions. 

(1) What is really responsible for the spontaneous inflammability of ordinary 
phosphine when brought in contact with air (R 461) ? 

(2) Compare the relative activities of phosphine and ammonia as reducing 
agents. 

Experiment XXXIII. 

The Phosphoric Acids. 
Discussion (R 465-468). 
Directions. 

(a) Metaphosphoric Acid. Throw 2 grams of phosphorus pentoxide in minute 
portions at a time into 10 c. cm. of water (?) in a small beaker, and allow 
the solution to stand for a few minutes, or until clear (Ei). Add a few 
drops of this solution to (1) a solution of albumen (?), (2) 5 c. cm. of a 
solution of ammonium molybdate and warm gently. Dilute a few drops 
with water, and add a solution of silver nitrate, a little at a time (E 2 ). 

63 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(6) Orthophosphoric Acid. Dilute the remainder of the solution of meta- 
phosphoric acid obtained in (a) with twice its volume of water; add a few 
drops of dilute nitric acid, and heat in the steam bath for half-an-hour (E 3 ). 
Repeat the tests made in (a) (E 4 -E 5 ). 

(c) Pyrophosphoric Acid. In a loop on the end of a platinum wire fuse 
to a clear melt enough disodium hydrogen phosphate to fill the loop (E 6 ). 
Boil the wire with its bead in a little water in a test-tube. Treat one 
portion of the resulting solution with silver nitrate (E 7 ) , and another with 
the solution of ammonum molybdate as in (a). Prepare a second bead 
of sodium pyrophosphate, dissolve it in 1-2 c. cm. of water, and test this 
solution with albumen solution as in (a) . 

(d) Hydrolysis of Orthophosphates. 

(1) Test the action on litmus paper of solutions of both of the acid 
orthophosphates of sodium. Explain the observed behavior, writing 
equilibria equations. 

(2) To 2 c. cm. of a solution of disodium hydrogen phosphate add silver 
nitrate solution until precipitation is complete (E 8 ). Test the resulting 
mixture with litmus paper. Explain, writing equilibria equations. 

(3) Pour off the supernatant liquid from the silver phosphate precipitate 
and add 2-3 c. cm. of a sodium or potassium chloride solution (?) (E 9 ). 
Test the resulting mixture with litmus paper. Explain, writing equilibria 
equations. 

Outside Questions. 

(1) Write graphic formulas for the three phosphoric acids. 

(2) Write equations EI, E 3 , and E 6 using graphic formulas. 

(3) Write graphic formulas for primary calcium phosphate (" superphos- 
phate ") 

(4) Predict what products will be formed on heating (A) sodium dihydrogen 
phosphate; (B) sodium ammonium hydrogen phosphate (" microcosmic 
salt")- 



Experiment XXXII. 
Phosphine. 

LABORATORY NOTES. 

Name Section ... 

/ 
Date ... Desk No. 



Experiment XXXIII. 
The Phosphoric Acids. * 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XXXIV. 
Phosphorous Acid. 

Discussion. Phosphorous acid exhibits a number of unusual and unex- 
pected properties. It is obtained by treating phosphorus trioxide with cold water. 
By analogy with nitrous acid, which may be obtained similarly from nitrogen 
trioxide, it would be expected to break up again, on heating, into the trioxide and 
water. One would also expect it to be an oxidizing agent. Moreover, by analogy 
with phosphoric acid, it should give three series of salts: Na 3 PO 3 , Na 2 HP0 3 , and 
NaH 2 PO 3 for instance. None of these expectations are realized. On heating, it 
decomposes into metaphosphoric acid and phosphine; it is a reducing rather than 
an oxidizing agent; only two of its hydrogen atoms ionize, and so only two types 
of salts can be prepared (NaH 2 P0 3 and Na 2 HPO 3 ). 

All of these peculiarities are, however, easily understood if we imagine that, 
in contrast with the phosphoric acids, one of the hydrogen atoms of phosphorous 
acid is directly attached to the phosphorus atom rather than to one of the oxygen 
atoms. In accordance with this view the structural or graphic formula would 
then be 

/O H /O H 

= P H instead of P O H 

\H \0 H. 

In this formula phosphine is already partly formed, and its production on 
heating phosphorous acid is therefore not surprising. Moreover, since phosphine 
is a strong reducing agent, it is not strange that phosphorous acid exhibits this 
same property. Finally, from our knowledge of phosphine, we can be reasonably 
confident that hydrogen directly united with a phosphorus atom will not ionize 
readily, and hence that only two series of salts of phosphorous acid can be formed. 
These deductions are evidently in the closest accord with the facts, and this lends 
a high degree of probability to the correctness of the formula proposed. 

Directions. Place a few drops of phosphorus trichloride in a dry test-tube 
and add 1-2 c. cm. of water (?). Warm (Ei), and test the vapor both with moist 
litmus paper and by bringing an open bottle of ammonium hydroxide solution 
near the mouth of the test-tube. What class of chlorides always reacts in this 
way with water ? Add an equal volume of molar cupric sulphate solution, and 
heat to boiling (?) (E 2 ). Add about a cubic centimeter of concentrated nitric 
acid; heat again to boiling (E 3 ) and cool. Interpret your results. 



65 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Outside Questions. 

(1) From analogy with the phosphoric acids predict what will be formed by 
careful dehydration of phosphorous acid ? Give names to these compounds 
and predict their properties. Make similar predictions regarding the 
effects of heating sodium acid phosphite. 

(2) What do the relative properties of ammonia and phosphine and of nitrous 
and phosphorous acids indicate regarding the relative oxidizing powers of 
nitric and phosphoric acids ? Cite other evidence derived from your 
experimental work in Chemistry 1 which confirms this conclusion. 



G6 



Experiment XXXIV. 
Phosphorous Acid. 

LABOEATORY NOTES. 

v 

Nam e - Section ... 

Date Desk No. 



Experiment XXXV. 

Carbon Monoxide. 
Discussion. (R 485-486).. 

Directions. Construct a small generator for the production of carbon 
monoxide from oxalic acid. This may be done conveniently as follows. Fit a 
safety tube and a short, right-angle delivery tube into the mouth of a 25 mm. 
test-tube. Attach to the delivery tube a right-angle extension tube long enough 
to reach to the bottom of another test-tube. Support the generator on a retort 
stand under the hood by means of a small iron clamp, and charge into it 2-3 
grams of oxalic acid and 5-10 c. cm. of concentrated sulphuric acid. Heat gently 
(Ei), and pass the resultant gas through a little lime or baryta (barium hydroxide) 
water (E 2 ). Be careful not to inhale this gas. It is poisonous. 

Attach a U-tube filled with soda-lime to the delivery tube from the gen- 
erator and support it by means of a second small clamp. Attach the extension 
tube to the outlet of the U-tube. Heat the reaction mixture again, if neces- 
sary, to produce a slow stream of gas, and test the gas as before with lime or 
baryta water. Coat the bottom, inside, surface of a narrow test-tube with 
copper oxide by the method used in Experiment XXX (6). Insert the exten- 
sion tube into this test-tube, and while a steady current of gas is maintained, 
heat the copper oxide to a red heat (E 3 ). 

Disconnect the extension tube, heat the reaction mixture sufficiently to 
produce a rapid current of gas, and apply a flame to the outlet of the U-tube (E 4 ) . 

Outside Questions. 

(1) Enumerate the properties of carbon monoxide disclosed by the above 
experiment. 

(2) Why was it advantageous to purify the gas ? 

Experiment XXXVI. 
Saturated and Unsaturated Hydrocarbons. 
Discussion (R 490-497). 
Directions. 

(a) To 5 c. cm. of gasoline in a dry test-tube add carefully, under the hood, a 
few drops of bromine [Caution!]. Breathe out across the mouth of the 
test-tube, and insert a piece of moist litmus paper into the tube. What 
gaseous substance is evidently formed ? 

(6) Fill the generator used in Experiment XXXV one-third full of alcohol, 
and dilute this with an equal quantity of water. Arrange the extension 

67 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

tube so that the gas escapes at the very bottom of a narrow test-tube con- 
taining two or three drops of bromine in a little water. Drop into the 
generator a few fragments of calcium carbide (Ei), and allow the gas to 
bubble through the solution for fifteen minutes, without attention ( ?) 
(E 2 ). When the bromine color has wholly disappeared, test the water for 
acidity. How does the action of bromine on acetylene differ from that on 
gasoline ? What does this show regarding acetylene ? 

(c) Apply a flame to a jet of the gas (E 8 ). 

(d) Optional. Mix thoroughly one drop of molar copper sulphate solution 
with about 250 c. cm. of water. To 5 c. cm. of this solution add a few 
drops of ammonium hydroxide solution (?) followed by a few drops of a 
solution of hydroxylamine sulphate, and warm. This reduces the am- 
monio cupric sulphate to ammonio cuprous sulphate. Pass acetylene 
gas from your generator through this solution (E 4 ). Do not allow the 
copper acetylide to dry, as it is highly explosive. 

Outside Questions. 

(1) What are the components of gasoline ? To what series do they belong ? 
What is the general formula for this series ? 

(2) Why was alcohol mixed with the water in generating the acetylene ? 



68 



Experiment XXXV. 
Carbon Monoxide. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XXXVI. 
Saturated and Unsaturated Hydrocarbons. 

LABORATORY NOTES. 

Name Section .... 

Date Desk No. 



Experiment XXXVII. 
Alcohols. 

Discussion. An alcohol may be defined as a compound containing one or 
more hydroxyl groups or radicals ( O H), each attached to a single carbon 
atom which in turn is attached to either carbon or hydrogen atoms. The follow- 
ing structural formulas of three important alcohols will illustrate this definition. 

H 
H H H H-C-O-H 

H-C-O-H H-C-C-0-H H-C-O-H 

i ii I 

H H H H-C-O-H 

I 
H 

Methyl, or Wood Ethyl, or Grain Propenyl Alcohol, 

Alcohol Alcohol or Glycerine. 

Alcohols may also be looked upon either as hydrocarbons in which one of the 
hydrogen atoms on each of one or more carbon atoms has been replaced by the 
hydroxyl radical, or as water in which one of the hydrogen atoms has been re- 
placed by a hydrocarbon radical. 

Alcohols derived from the simpler hydrocarbons resemble water much more 
than they do the hydrocarbons. Thus the hydrogen of their hydroxyl group can 
be replaced by an active metal just as it can in water. Potassium, for instance, 
reacts vigorously with methyl alcohol to form potassium methyl alcoholate and 
hydrogen ; 

2K + 2CH 3 OH = 2CH 3 K + H 2 . 
With water the corresponding reaction would be, of course, 

2K + 2HOH = 2HOK + H 2 . 

Alcohols derived from the higher hydrocarbons, that is, from those contain- 
ing many carbon atoms, exhibit these similarities to water to a much less pro- 
nounced degree. Indeed, they resemble more closely the hydrocarbons from 
which they are derived. 

Alcohols are in general reducing agents, both in the gaseous condition and 
in aqueous solution, though this property varies greatly among different members 
of the group. 

69 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Finally, alcohols resemble inorganic bases in that with acids they form com- 
pounds analogous to salts. Thus, ethyl alcohol and hydrochloric acid react to 
form ethyl chloride and water; 

C 2 H 5 OH + HC1 C 2 H 5 C1 + H 2 0. 

This reaction is reversible, and so with dilute hydrochloric acid is very incomplete; 
but if concentrated acid is used and if some provision is made for absorbing the 
water which is formed in the reaction, such as by adding a hygroscopic sub- 
stance like zinc chloride, it can be made to go nearly to completion. These 
reactions between alcohols and acids, however, differ strikingly in speed from 
the analogous reactions between inorganic bases and acids. They usually require 
hours for completion, even at. high temperatures, while neutralization reactions 
are apparently instantaneous. This difference is doubtless connected with the 
fact that alcohols are very slightly, if at all, ionized; and that the reaction with 
acids is therefore one between the undissociated molecules of the alcohol and the 
acid, rather than between the hydroxyl and hydrogen ions, as in neutralization. 

The compounds formed when alcohols and acids react are called esters 
instead of salts. This serves to emphasize both that they are not formed by a 
true neutralization reaction, and that they are not ionized in aqueous solutions. 
Consequently, the formation of esters from alcohol and acids is called esterifica- 
tion. The reverse action, that is, the formation of an acid and an alcohol by 
the action of water on an ester, 

C 2 H 5 C1 + H 2 O = C 2 H 5 OH + HC1, 
or the action of an inorganic base on an ester to form a salt and an alcohol, 

CH 3 COOC 2 H 5 + NaOH = CH 3 COONa + C 2 H 5 OH, 
is called saponification (R 505). 

Directions. 

(a) Observe the odor of ethyl alcohol. Put a drop or two on the tongue [care 
- poison !] Test a little of it with litmus paper. Is alcohol an acid or a 
base ? 

(6) Heat a narrow strip of copper gauze in a flame until it is covered with a 
layer of oxide. Place a few drops of methyl alcohol in a test-tube, heat 
until all the alcohol has vaporized, then plunge the hot, oxidized copper 
gauze into the test-tube. What property of alcohol does this illustrate ? 

(c) To a few cubic centimeters of absolute methyl or ethyl alcohol in a dry 
test-tube add a short length of sodium wire (Ei). Collect a little of the 
resulting gas over water, and convince yourself as to its nature. Is the 
action more or less vigorous than with water ? 

70 



EXPERIMENT XXXVII 

(d) To 5 c. cm. of dilute sodium hydroxide solution add a crystal of iodine and 
two or three drops of alcohol. Warm slightly. Note the odor. The pro- 
duction of iodoform in this way is a test for ethyl alcohol. Unfortunately 
there are several other substances which behave similarly, so that the 
test is not wholly conclusive. 

Outside Questions. 

(1) From analogy with inorganic hydroxides and oxides, write the structural 
and empirical formulas of (A) methyl oxide; (J5) ethyl oxide. What are 
such compounds called (R 506) ? 

(2) Write the structural and empirical formulas of (A) the methyl ester of 
sulphuric acid; (J5) the glycerine ester of hydrobromic acid. 



Experiment XXXVIII. 

Organic Acids. 
Discussion. An organic acid may be defined as a compound containing the 

carboxyl group I C / J attached to a hydrogen or a carbon atom. The 

following structural formulas of important organic acids will illustrate this 



definition : 



O 
// 



H O 



O H C C O H C H 

n I ^ 

H-C H H 

Formic Acid Acetic Acid Oxalic Acid. 

Organic acids in general are very much weaker, that is, much less dissociated, 
than the strong inorganic acids. As for their other properties, no general state- 
ment can be made, for among the many thousands of these substances which have 
been prepared almost every sort of physical and chemical property finds exempli- 
fication. Brilliant dyes, pleasant perfumes, evil odors, corrosive poisons, and 
beneficial foods, all are represented. 

Organic acids may be prepared in many ways, but the most general and 
important is by the oxidation of alcohols. Thus, methyl alcohol when oxidized 
yields formic acid, according to the equation, 

H H o 

H C H + 20 = H C H = H C H + H 2 
l I 

H H 

Methyl Alcohol Formic Acid. 

71 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Directions. 

(a) To one gram of sodium acetate, add a little dilute sulphuric acid, and 

warm (Ei). Note the odor. 

(6) Add a few drops of glacial acetic acid to 1-2 c. cm. of ethyl alcohol. Add 
1-2 c. cm. of concentrated sulphuric acid, warm gently (E 2 ), and note the 
odor. This serves as a characteristic test for acetic acid. 

(c) To a few cubic centimeters of sodium hydroxide solution in a test-tube 
add a few drops of ethyl acetate. Insert a slightly smaller test-tube, 
filled with cold water, well into the mouth of the first test-tube. Heat 
the mixture, at first gradually, but later to the boiling point, until the odor 
of ethyl acetate can no longer be detected (E 3 ). The smaller test-tube, 
filled with cold water, will prevent the loss of volatile substances. Refill 
it with cold water if necessary. Finally cool and apply the iodoform 
test for ethyl alcohol as in Experiment XXXVII. 

(d) Dissolve a few crystals of potassium permanganate in 5 c. cm. of dilute 
(3M) sulphuric acid by warming. Add 3^ drops of ethyl alcohol, and 
heat. If, after a few minutes, the hydrated manganese dioxide has not 
dissolved, cool, and then add (danger!} an equal volume of concentrated 
sulphuric acid, and heat further (E 4 ) . When colorless, cool and add a cubic 
centimeter of alcohol, warm gently, and note the odor (E 5 ). What sub- 
stance must have been present after the treatment with permanganic acid? 

(e) Add 10 c. cm. of alcohol to 4-5 grams of cottonseed oil in a porcelain 
dish. To the resulting mixture add about 1 c. cm. of a 50 percent aqueous 
solution of sodium hydroxide. Boil very gently with constant stirring, 
until the odor of alcohol is no longer perceptible (E a ). In writing the 
equation assume that the oil consists wholly of tripalmitin (R 505) ; the 
alcohol is used merely as a common solvent for the oil and the sodium 
hydroxide. What remains in the dish ? 

Dissolve the soap in 50 c. cm. of warm water, and filter. Cool the 
filtrate, and to half of it add dilute hydrochloric acid (E 7 ), and shake 
vigorously ( ?). Withdraw the floating semi-solid substance with a glass 
rod, transfer it to a little sodium hydroxide solution and warm (E 8 ). Add 
small portions of the remaining filtrate to test-tubes containing dilute 
solutions of calcium sulphate, magnesium sulphate, and sodium sulphate, 
respectively (E 9 _ n ). Explain your results. 

Outside Questions. 

(1) What was the function of the sulphuric acid in (6) ? Write an equilibrium 
equation showing it. 

(2) What does (6) show regarding the relative properties of the ethyl esters 
of acetic and sulphuric acids ? 

(3) What difference between esters and salts is brought out by (c) ? What 
other marked differences are there ? 

72 



Experiment XXXVII. 
Alcohols. 

LABORATOEY NOTES. 

Name Section .. 



Date Desk No. 



Experiment XXXVIII. 
Organic Acids. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XXXIX. 

Boron and Silicon. 
Assigned Reading. Smith, pp. 518-529. 

Discussion. In Mendelejeff's periodic table of the elements boron comes 
next on the left to carbon, while silicon comes just below carbon in the same 
group. It is not surprising, therefore, that they both resemble carbon. More- 
over, since they both differ from carbon in similar ways, they resemble each 
other even more markedly. Both elements are but slightly less hard than the 
diamond, and but slightly less refractory than carbon. Both form gaseous hy- 
drides analogous to the hydrocarbons, as well as compounds with hydrogen and 
the halogens surprisingly similar to the corresponding compounds of carbon, such 
as chloroform, carbon tetrachloride, etc. 

Directions. 

(a) Borates and Boric Acid. 

(1) Dissolve 5 grams of borax in 15 c. cm. of boiling water. Test the solu- 
tion with litmus paper (Ei). Add 5 c. cm. of concentrated hydrochloric 
acid to the hot solution (E 2 ), and cool under the tap. Filter off the 
crystals, and wash them with a little cold water. 

(2) Dry some of the crystals on a piece of nickel or platinum foil or in an 
iron dish or spoon, and fuse (E 3 ). 

(3) Add some of the crystals to a mixture of 10 c. cm. of methyl alcohol 
and 5 c. cm. of concentrated sulphuric acid in a casserole (?). Set fire to 
the mixture and stir with a glass rod (E 4 ) . 

(4) Recrystallize the remainder of the boric acid from a little hot water, 
wash as before, and dissolve the resulting crystals in hot water. Test this 
solution with litmus paper. Dip a piece of turmeric paper in this solution; 
wrap it around the upper part of the text-tube and boil the solution until 
the paper is dry. Touch the paper with a drop of sodium hydroxide 
solution. This is a test for boric acid. 

(b) Silicates and Silicic Acid. Mix about a half a gram of sand with 3-4 
grams of sodium hydroxide powder in a clean, dry, iron crucible. Heat 
gently over a Bunsen flame until effervescence (?) ceases, and then to a 
red heat until solidification occurs (E 5 ). Cool thoroughly, dissolve the 
product as completely as possible in 15-20 c. cm. of hot water, cool, and 
filter the solution from the hydroxides of iron and from any sand which 
may have escaped attack. Acidify the filtrate with concentrated hydro- 
chloric acid (Ee), transfer the mixture to a casserole, evaporate to dryness 

73 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

under the hood, and finally ignite in the bare flame (E 7 ). Cool, extract 
the sodium chloride from the residue with water, and identify what 
remains. 

(c) Solubility of Glass. Carefully clean a broken test-tube, crush it in a wet 
mortar and grind it to a coarse powder. Avoid the danger of flying 
particles of glass during the crushing by placing over the mortar a piece 
of cloth or paper with a hole in it for the handle of the pestle. Rinse the 
powder in the mortar several times with distilled water, finally covering 
the powder with water. Add a few drops of phenolphthalein solution and 
grind the powder further (E 8 ) (R 606). 

(d) The Chemist's Garden. Mix thoroughly in a small beaker 30 c. cm. of 
the commercial solution of sodium silicate known as " water glass," with 
about twice its volume of water. Drop a small crystal of each of the 
following substances into this solution: copper sulphate, nickel sulphate, 
cobalt nitrate, zinc sulphate, ferrous sulphate. Observe the behavior of 
these crystals. If growth is still occurring at the end of the laboratory 
period, cover the beaker and place it with its contents where it can remain 
undisturbed until the next period. 

Outside Questions. 

(1) Explain the preparation of boric acid, writing equilibria equations. 

(2) What was the purpose of the sulphuric acid used in (a-3)? See Experi- 
ment XXXV and (R 504). 

(3) Explain the changes observed in (d) (R 283). 



74 



Experiment XXXIX. 
Boron and Silicon. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XL. 
Alkaline Earth Elements. 

Discussion. Dobereiner in 1829 pointed out that the majority of the ele- 
ments then known could be arranged in groups or families of three, all three 
members of each family exhibiting pronounced similarities. Of these "triads," 
chlorine, bromine, and iodine; lithium, sodium, and potassium; iron, nickel, and 
cobalt; and calcium, strontium, and barium were striking examples. Now, we 
recognize that this regularity is only an imperfect one, and is included in the 
much more far-reaching Periodic System of the elements. 

The following experiments are designed to bring out some of the differences, 
as well as some of similarities, of the members of the alkaline-earth triad. 

Directions. 

(a) Treat solutions of any soluble salts of calcium, strontium, and barium with 
a solution of sodium carbonate (Ei). Repeat, using a solution of sodium 
hydrogen carbonate (sodium bicarbonate); warm (E 2 ). What do you 
conclude regarding the solubilities of the neutral and acid carbonates of 
this group ? 

(6) Add a solution of ammonium oxalate to solutions of any soluble salts of 
calcium, strontium, and barium (E 8 ). Add dilute hydrochloric acid. 
Explain, illustrating your explanation with an equilibrium equation 
(R 598-601). What do you conclude regarding the solubilities of the 
oxalates of this group ? 

(c) Add a solution of strontium chloride to a saturated solution of calcium 
sulphate. Add a solution of barium chloride to a saturated solution of 
strontium sulphate. What is shown regarding the relative solubilities 
of the sulphates of this group ? 

(d) Repeat (c) using saturated solutions of calcium and strontium chromates 
instead of the corresponding sulphate solutions. What is shown regard- 
ing the relative solubilities of the chromates of this group ? 

Outside Questions. 

(1) Explain on the basis of your observations in (a) how limestone and dolo- 
mite rocks are dissolved, and how stalactites and similar formations are 
produced. 

(2) To distinguish (A) between barium and strontium, and (B) between stron- 
tium and calcium, would the chromate or the sulphate method be prefer- 
able (R 544) ? 



75 



Experiment XL. 
Alkaline Earth Elements. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XLI. 

Magnesium. 

Discussion. -Magnesium may be looked upon as a transition element be- 
tween the alkaline earth elements on the one hand, and the elements of the zinc 
group (zinc, cadmium, and mercury) on the other. Its position in the displace- 
ment series, the strength of its hydroxide as a base, and the hydrolysis and 
solubilities of its salts, all justify this point of view. 

Directions. 

(a) Test a solution of magnesium chloride with litmus paper. Heat gradu- 
ally (?) to dryness some crystals of this salt (formula ?) in a dry test-tube 
EO. Test with litmus paper the liquid which condenses on the walls of 
the tube; then remove this liquid by warming, or with a piece of filter 
paper, and test the solubility of the residue in water, and the action of 
the resulting mixture upon litmus paper. Explain what you observe. 
What do you conclude regarding the strength of magnesium hydroxide 
as a base ? 

(6) Add a solution of sodium hydroxide, a little at a time until in excess, to a 
solution of any soluble magnesium salt (E 2 ). Add an excess of ammonium 
chloride solution to the resulting mixture (E 3 ) . Explain this phenomenon, 
and write equilibria equations illustrating your explanation (R 644). 

(c) Clean a piece of magnesium ribbon with a little dilute acid; rinse off the 
acid, and cover the ribbon with water. Add a little ammonium chloride 
solution to the water (E 4 ). Explain this phenomenon with the aid of 
your observations and conclusions in (6). 

(d) Mix thoroughly in a mortar about a cubic centimeter each of magnesium 
powder and powdered calcium carbonate. Put the mixture in a dry test- 
tube, support the tube in a clamp on a retort stand, and heat the top 
layer in the Bunsen flame until the reaction begins. Be careful to keep 
the tube directed away from your face during the heating. (Danger !) 
Allow the test-tube to cool, add a little water, and then, slowly, an excess 
of concentrated hydrochloric acid (?) . (If the tube has been broken, place 
the contents with the acid in a beaker.) The acid will dissolve the oxides 
of magnesium and calcium formed by the action. What effect will the 
acid have upon any excess of either of the ingredients originally present ? 
When all action has ceased, filter and wash the black residue (?) with 
water. After drying this on the water-bath, prove that it is carbon. To 



77 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

do this place some of it in a dry test-tube, add a pinch of potassium 
chlorate, heat in the Bunsen flame, cool (keeping the tube closed with 
the thumb while waiting for this), and pour the gas when cooled into a 
test-tube containing two cubic centimeters of lime-water, and shake (?). 
What does this experiment demonstrate regarding the relative reducing 
powers of carbon and magnesium ? How, therefore, must magnesium 
compare in this respect with most of the other elements ? 

Outside Questions. 

(1) Explain why sea water cannot be used in steam boilers. 

(2) What would you conclude from (6) regarding the solubility of magnesium 
hydroxide in water ? Arrange zinc, cadmium, mercuric, calcium, stron- 
tium, and barium, hydroxides in the order of their solubilities (R 544) . 
The solubilities of cadmium and mercuric hydroxides are 2.6XlO~ 4 and 
2.4X10~ 4 molar, respectively. 



78 



Experiment XLI. 
Magnesium. 

LABORATORY NOTES. 

Name ; Section ... 

Date ... Desk No. 



Experiment XLII. 

Zinc, Cadmium, and Mercury. 

Discussion. In this sub-group, as in corresponding sub-groups of the 
periodic table, the metals become nobler and also more prone to form complex 
ions, as their atomic weights increase. There is, however, no regular gradation 
in the degrees of dissociation of the hydroxides of these elements. 

Directions. 

(a) Test a solution of zinc sulphate with litmus paper, and then add carefully, 
a drop at a time, a dilute solution of sodium hydroxide (Ei-E 2 ). Explain 
what each of your observations indicates regarding zinc hydroxide. Illus- 
trate your answer with equilibria equations. 

(6) Into the clear solution obtained from (a) pass hydrogen sulphide gas (E 3 ). 
Make slightly acid with hydrochloric acid (E 4 ) . Add an excess of sodium 
acetate solution. Explain, writing illustrative equilibria equations. 

(c) To about 5 c. cm. of zinc sulphate solution add sodium carbonate solution, 
at first a little, and then in excess (?) (E 5 ). Bring the contents of the 
tube to the boiling point, filter, and wash the precipitate with water. 
To a portion of the precipitate add an acid (E 6 ). 

(d) To a little zinc dust in a dry test-tube add concentrated sodium hydroxide 
solution and warm (E 7 ). Identify the gas evolved. 

(e) Add a drop of cobalt nitrate solution to a cubic centimeter of water, and 
with two or three drops of this solution moisten a little zinc oxide in a 
dry test-tube. Warm the moist zinc oxide until all the water has been 
driven off, and then ignite strongly (E 8 ) . The resulting cobalt zincate is 
one of the very few colored compounds of zinc, and is, therefore, an im- 
portant test. 

(/) Test a solution of cadmium sulphate with litmus paper, then add sodium 
hydroxide solution in excess (E 9 ) . What do you infer from these observa- 
tions regarding cadmium hydroxide ? 

(g) Pass hydrogen sulphide into a solution of cadmium sulphate containing a 
little sulphuric acid (Ei ). 

(h) To dilute solutions of mercurous nitrate and of mercuric nitrate add 
separately the following reagents : (1) litmus paper ; is mercurous or 

79 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

mercuric hydroxide the stronger base ? how does mercuric hydroxide 
compare with cadmium hydroxide in this respect ? (2) dilute hydro- 
chloric acid ; (3) sodium hydroxide ; (4) hydrogen sulphide (En-Ei 4 ) 
(R 656-657) ; (5) potassium iodide solution until there is no further 
change; what complex ion is formed (R 656) ? Show by experiment 
that mercuric sulphide gives an even smaller concentration of mercuric 
ions than does this complex ion. 



80 



Experiment XLII. 

Zinc, Cadmium, and Mercury. 

LABORATORY NOTES. 

Name Section 

Date ... Desk No. 



Experiment XLIII. 

Overvoltage. 

Discussion. In the discussion of the displacement series of the metals 
Experiment XXVI) it was seen that, since the displacement of one metal from 
a solution of another metal involved merely the transfer of positive electricity 
from the atoms of one metal to those of the other, this series must also represent 
the relative attractions of the atoms of the different metals for positive electricity. 
Hydrogen occupies an intermediate position in this series. It is displaced 
from its solutions (the acids) by metals above it, such as zinc and iron, but not 
by metals below it, such as copper and mercury. Hydrogen, however, exhibits 
certain peculiarities in this respect, which may be connected with the fact that 
; is a gas, while the other elements which acquire positive charges in solution 
are solids. Thus hydrogen is not evolved by pure zinc present alone in an acid 
solution; the zinc must contain certain metallic impurities, or it must be in con- 
tact with certain other metals, if hydrogen is to be evolved at all rapidly. That is, 
hydrogen exhibits a peculiar reluctance to appear as a gas on the surface of 
many metals. This reluctance, which is different for different metals, is called 
supertension, or overvoltage. The following experiment will serve to illustrate 
this property. 

Directions. 

(a) Drop a piece of pure granulated zinc into dilute sulphuric acid. Touch 
the zinc with a piece of platinum wire (?). Explain with a diagram the 
changes which occur at the surfaces of the platinum and the zinc, and 
the path followed by the positive electricity. Why must the wire touch 
the zinc ? 

(6) Add a drop of copper sulphate solution to the zinc and sulphuric acid of 
(a), and shake (?). Explain all that you observe. Is the overvoltage of 
hydrogen on copper greater or less than on zinc ? 

(c) Add a few drops of mercuric nitrate solution to the mixture obtained in 
(6) (?). Again touch the zinc with the platinum wire, leaving the wire in 
contact for some time (?). Explain all that you observe. Arrange the 
various metals studied in the order of decreasing overvoltage toward 
hydrogen. 



81 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Outside Questions. 

(1) Do you know any other property of gases which resembles the over- 
voltage exhibited by hydrogen on various metals ? 

(2) What would you predict would be the effect of adding a little cobalt 
nitrate solution to dilute sulphuric acid in which a piece of pure zinc is 
immersed, knowing that the overvoltage of hydrogen on cobalt is less 
than on zinc ? What would you predict would be the effect of adding 
mercuric nitrate solution to this mixture, knowing that mercury does 
not form an amalgam with cobalt ? 



82 



Experiment XLIII. 
Overvoltage. 

LABORATORY NOTES. 

Name : Section ... 

Date ... Desk No. 



Experiment XLIV. 

Aluminum. 
Discussion (R 681-691). 

Directions. 

(a) Clean some aluminum chips with a little dilute hydrochloric acid, rinse 
them thoroughly with water, and then observe whether they act upon 
water. Place one of the chips in contact with a piece of platinum wire. 
Then add a few drops of mercuric chloride solution (Ei), shake, rinse with 
water, and again examine the behavior toward water (?) (E 2 ). Explain. 

(6) Compare the actions of dilute nitric and dilute sulphuric acids with that 
of dilute hydrochloric acid on aluminum chips. Explain on the basis of 
your conclusion in (a). What objection is there to this explanation ? 

(c) Heat some aluminum chips with concentrated sodium hydroxide solution 
(E 3 ). Carefully neutralize a portion of the solution with dilute hydro- 
chloric acid (E 4 ). To the remainder add barium chloride solution (E 6 ). 

(d) Test a solution of aluminum sulphate with litmus paper. What do you 
infer ? Add sodium hydroxide solution, a little at a time, but finally in 
excess (E 6 -E 7 ). What does this experiment demonstrate ? 

(e) To a solution of aluminum sulphate add a solution of sodium carbonate 
(E 8 ). Filter off the precipitate, wash it until free from sodium carbonate, 
and then ascertain whether it is a carbonate or not. 

(/) Repeat (e} using a solution of ammonium sulphide in place of the sodium 
carbonate (E 9 ). Ascertain whether the precipitate is a sulphide or not. 

(g) To a few drops of aluminum sulphate solution add a drop of cobalt 
nitrate solution. Soak a small piece of filter paper in this liquid, roll it up 
tightly and wind the end of a platinum wire several times about it. 
Dry the paper over a flame, burn it, and finally ignite the ash thor- 
oughly. The resulting cobalt aluminate is known as Thenard's Blue, 
Co(A10 2 ) 2 (E 10 -E). 

(h) Boil separate pieces of cotton cloth in water, and in solutions of aluminum 
sulphate, ferric chloride, and sodium stannate. Squeeze out the super- 
fluous liquid, and boil each piece for a few minutes in a solution of alizarin 
made slightly ammoniacal, and then rinse under the tap. Explain your 
observations (R 689). 

(i) Warm 5 c. cm. of saturated ammonium sulphate solution, and add to 
it an equal volume of a saturated aluminum sulphate solution (E 13 ). 

83 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

Optional. If you wish to obtain some large crystals of this alum, saturate 
10 c. cm. of warm water with aluminum sulphate, and add it to an equal 
volume of a hot saturated solution of ammonium sulphate in a small 
beaker. Hang a thread in the warm solution, and allow the solution to 
cool slowly. 

Outside Questions. 

(1) Explain your conclusions in (e) and (/). 

(2) To what class of minerals does cobalt aluminate correspond (R 686) ? 

(3) Write the formula for potassium chrom-alum. 



84 



Experiment XLIV. 
Aluminum. 

LABORATORY NOTES. 

Name Section ... 

Date ... l^esk No. 



Experiment XLV. 
Arsenic, Antimony, and Bismuth. 
Assigned Reading. Smith, pp. 707-720. 

Discussion. The sub-group, arsenic, antimony, and bismuth, following 
nitrogen and phosphorus in the fifth group of the periodic table, illustrates in a 
striking fashion the gradual transition, within a single group, from a pronounced 
non-metallic element at the top of the periodic table to a pronounced metallic 
element at the bottom. 

Solid nitrogen does not exhibit the slightest metallic property, nor does 
phosphorus in its ordinary modifications. Its black modification, discovered by 
Bridgman, does, however, show a metallic lustre and some ability to conduct 
heat and electricity. The ordinary, stable modifications of arsenic and antimony 
are metallic ; but the yellow, non-metallic modification is unstable and ephemeral 
except at very low temperatures. In the case of bismuth no non-metallic modifi- 
cation has yet been prepared. 

Directions. 

(a) Preparation and Properties of Arsine and Stibine. Arrange a small flask 
with a safety tube and a straight delivery tube provided with a horizontal 
nozzle 15 cm. long. Place in the flask a piece of chemically pure zinc, and 
add pure (?) hydrochloric acid. When the air has been displaced (Care! 
Test ?), light the jet of gas and hold a porcelain crucible-lid in the flame. 
If there is a deposit, use purer zinc and acid. Otherwise, add a few 
drops of a solution of arsenic trichloride (Ei). Observe the appearance 
of the flame, and hold the crucible-lid in it (E 2 ). Heat the middle of 
the nozzle-tube with a colorless flame (Marsh's Test) (E 3 .) Apply some 
freshly prepared bleaching powder solution to the deposit on the lid (E 4 ). 
To prevent the escape of arsine, do not put out the hydrogen flame until 
these experiments with arsenic are completed. When they are, however, 
fill the generator with water, rinse it out, attach a fresh nozzle-tube, 
charge with fresh zinc and acid, and repeat the experiment, using a few 
drops of a solution of antimony trichloride in place of the one of arsenic 
trichloride (Es-s). 

(6) Hydroxides. Dilute one cubic centimeter each of 2 M solutions of arsenic, 
antimony, and bismuth chlorides with 20 c. cm. of water (?) (E 9 ). Clarify 
if necessary with dilute hydrochloric acid. Divide each solution into three 

85 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

portions. To one portion of each solution add sodium hydroxide solution, 
at first gradually, and then in slight excess (E 10 _ 12 ). To the resulting mix- 
tures add a little starch solution, and then a few drops at a time of a 
0.5 M solution of iodine (E 1S _ 14 ) (R 712). 

(c) Sulphides and Sulphur Acids. Into another portion of each of the solu- 
tions prepared in (6) pass hydrogen sulphide gas until no further precipita- 
tion occurs (E 15 _ 17 ). Neutralize with a solution of ammonium hydroxide 
(E 18 . 19 ). If complete solution does not occur, add a little finely divided 
sulphur, and shake (E^a). Finally add hydrochloric acid, in excess, to 
each solution (E 22 . 24 ). How could you detect a small quantity of bismuth 
in a solution containing an antimony salt ? 

Outside Questions. 

(1) Enumerate the facts discovered in the above experiments which, in con- 
nection with what you already know about nitrogen and phosphorus, dis- 
close the gradual transitions in properties which occur in this group of the 
periodic table. 

(2) Write equilibria equations showing the electrolytic dissociation of the 
ammonium salts of the sulpho- acids of arsenic and antimony (R 713). 
By analogy with " anhydrides," what would you call the various sulphides 
of these metals ? Would you expect the sulpho- acids to be stronger or 
weaker than the corresponding oxygen acids ? Why ? 



86 



Experiment XLV. 
Arsenic, Antimony, and Bismuth. 

LABORATORY NOTES. 

Name Soctien ... 

Date ... Desk No. 



Experiment XLVI. 

Tin and Lead. 
Assigned Reading. Smith, pp. 692-705. 

Discussion. - - Tin and lead are in the same group of the periodic table as 
carbon, and in their tetravalent compounds they show decided similiarities 
to corresponding carbon compounds. This is especially true of tin. In the free 
state, however, they show pronounced metallic properties; indeed, they are. 
altogether different from carbon. In the divalent condition the metallic proper- 
ties predominate, and both tin and lead form stable salts of this series. 

Directions. 

(a) The Position of Tin in the Displacement Series. Place a strip of zinc in 
a solution of stannous chloride (Ei). Then place a piece of granulated 
tin in a solution of copper sulphate and warm (?). What are the relative 
positions of zinc, tin, and copper in the displacement series ? 
(6) Stannous Salts. 

(1) Dissolve half a gram of tin in warm, concentrated hydrochloric acid 
(E 2 ). (While solution is occurring, proceed with Experiment (c), below.) 

(2) Pour a few drops of this solution into a solution of hydrogen sulphide 

(Ea). 

(3) Dilute a little of the solution from (1), and add sodium hydroxide 
solution, a drop at a time, until in excess (E 4 -E 5 ). 

(4) Pour a little of the solution, a drop at a time, into a solution of mer- 
curic nitrate (Ee). 

(5) To 1-2 c. cm. of the solution, add bromine-water until the color 
ceases to be destroyed (E 7 ). Drive off the excess by warming (hood), and 
test for stannic ions according to (c). 

(c) Stannic Salts. 

(1) Add (caution!) (hood) two drops of stannic chloride to 20 c. cm. of 
water in a test-tube (E 8 ). 

(2) Pour a few drops of this solution into a solution of hydrogen sulphide 
(E 9 ). 

(3) Dilute a little of the solution, and add, a drop at a time, a solution of 
sodium hydroxide, until in excess (Ei -En). 

(4) Pour a few drops of the solution into mercuric nitrate solution (E). 

(d) Position of Lead in the Displacement Series. Place a strip of zinc in a 
solution of lead acetate. Place a strip of lead, or better a little lead wool, 
in a solution of copper acetate, and warm. What are the relative posi- 
tions of zinc, lead, and copper in the displacement series ? 

87 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(e) Plumbous Salts. Use diluted lead nitrate solution. 

(1) Add a solution of sodium hydroxide, a little at first, and then in 
excess (E 13 ). 

(2) Add dilute hydrochloric acid. Filter a part of the resulting solution, 
and add dilute sulphuric acid and hydrogen sulphide to separate portions 
of the filtrate. Heat the remainder of the unfiltered solution. What do 
you conclude regarding the relative solubilities of the chloride, sulphide, 
and sulphate of lead ? 

(/) Plumbic Compounds. 

(1) Lead Dioxide. Warm and agitate one gram of minium (red lead) 
with 5-6 c. cm. of dilute nitric acid until it no longer changes in color 
(Ei 4 ). Dilute with water and filter. Test for lead ions in the filtrate. 
Treat the precipitate with sodium hydroxide solution (Ei 5 ). 

(2) Ammonium hexachlor plumbate. Cool 10 c. cm. of concentrated 
hydrochloric acid, diluted with 3 c. cm. of water, in ice-water under the 
hood. To this add 10 grams of lead dioxide, and allow the whole to 
stand for ten minutes (Ei 6 ). Filter through an asbestos filter, keeping 
the filtrate cold. Meanwhile, prepare 8 c. cm. of a saturated solution 
of ammonium chloride in cold water, and add this to the filtrate (Ei 7 ). 
Filter off the precipitate, and wash with a little concentrated hydro- 
chloric acid. 

Dissolve a little of this salt in a small quantity of water, and divide 
the solution into two parts. Dilute one portion further with water and 
let stand (Ei 8 ). Warm the other portion (Ei 9 ). 

(3) Add cautiously, a little at a time, 5 grams of the hexachlor plumbate 
to 5 c. cm. of cold, concentrated sulphuric acid under the hood. Let 
stand for some time in ice-water (E 2 o). 

Outside Questions. 

(1) Explain all you observed when tin was heated with copper sulphate. 

(2) In (d) why was the acetate of copper employed rather than other common 
copper salts ? 

(3) What derivative of hexachlor plumbic acid corresponds to the anhydride 
of an oxygen acid ? 



Experiment XLVI. 
Tin and Lead. 

LABORATORY NOTES. 

Name Section ... 

Date ... Desk No. 



Experiment XLVII. 
Colloidal Solutions. 

Discussion. Solids and liquids can, under proper conditions, be produced in 
other liquids in such a fine state of subdivision that their particles never settle 
out, pass through all ordinary niters, and are invisible even with the most powerful 
microscopes. Nevertheless these particles are not really in solution, for some 
filters will remove them, the so-called ultra-miscroscope can detect them, and the 
liquids containing them do not behave, in several respects, like ordinary solutions. 
Substances in this condition of very fine subdivision are called colloids, and 
liquids containing these colloids are called colloidal solutions. Such solutions are 
evidently intermediate between ordinary suspensions of fine particles, and true 
solutions where the dissolved particles are of molecular size. 
Directions. 

(a) Preparation. Boil 3 grams of arsenic trioxide (white arsenic) with 25 c. cm. 
of distilled water in a small flask. Filter while hot, cool the filtrate to 
room temperature, and pass in hydrogen sulphide until the solution turns 
yellow. Blow out the excess of hydrogen sulphide with a stream of air 
from the compressed-air supply. When the solution no longer smells of 
hydrogen sulphide, add some of the arsenic trioxide previously filtered off, 
and boil again. Filter, cool and pass in hydrogen sulphide as before. 
Blow out the excess, and if, as is usually the case, some arsenic trisulphide 
now separates out, filter it off. The filtrate is a colloidal solution of 
arsenic trisulphide, and may be used for the following coagulation 
experiments. 

(6) Coagulation. 

(1) Place 2 c. cm. of the colloidal solution in each of four test-tubes. 
Heat the contents of one test-tube. To the second test-tube add 3 drops 
of sodium chloride solution, to the third three drops of barium chloride 
solution, and to the forth three drops of aluminum chloride solution, 
performing these operations simultaneously. 

(2) To another 2 c. cm. sample of the colloidal solution add an equal 
volume of gelatine solution (also a colloidal solution), and then three 
drops of barium chloride solution. 

(3) Add 10 drops of ammonium hydroxide solution to 50 c. cm. of dis- 
tilled water. Prepare a solution of ferric chloride so dilute that it has a 

89 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

pale yellow color. While constantly stirring the ammonium hydroxide 
solution, pour in, drop by drop, 1 c. cm. of the ferric chloride solution. 
Mix a few cubic centimeters of the resulting colloidal solution of ferric 
hydroxide with an equal volume of the colloidal solution of arsenic 
sulphide. 

Outside Questions. 

(1) When a colloidal solution of arsenic sulphide is electrolysed, the minute 
particles of sulphide move toward the positive electrode. Colloidal ferric 
hydroxide, when similarly treated, moves toward the negative electrode. 
Knowing these facts, devise an explanation for your observations in 
(6-2) and (6-3). 

(2) What sort of ions would you expect to coagulate colloidal solutions of 
ferric hydroxide most readily ? 



90 



Experiment XLVII. 
Colloidal Solutions. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



Experiment XLVIII. 

Manganese. 
Assigned Reading. Smith, pp. 737-745. 

Discussion. Manganese is the most versatile of the elements. It exhibits 
no less than five different valences in its various compounds, and its properties 
in these different stages of oxidation are strikingly different. The metal itself is 
almost as electropositive as aluminum, and so dissolves in acids with the greatest 
ease, even displacing zinc from solutions of zinc salts. In its divalent condi- 
tion it forms a relatively strong base, while in its heptavalent condition it forms 
an oxygen acid almost as strong as nitric or hydrochloric acid. Its divalent 
compounds are pink, its trivalent are violet, its tetravalent brown or black, its 
hexavalent green, and its heptavalent purple. It is no wonder that potassium 
manganate is called " mineral chameleon "! 

Directions. 

(a) Manganous Salts. 

(1) Test some manganous sulphate solution with litmus, and then add 
sodium hydroxide solution in excess (Ei). Save the resulting mixture 
for (c). What do you infer regarding manganous hydroxide ? 

(2) To a solution of a manganous salt add a solution of ammonium sul- 
phide. Ascertain whether the product is, or is not, a sulphide (E 2 ). 

(6) Manganic Salts. 

(1) To a little of a solution of manganous sulphate in a porcelain crucible 
add a few cubic centimeters of concentrated phosphoric acid, a few drops 
of concentrated nitric acid, and boil off a large part of the water (E 3 ). 
Then dilute with water (E 4 ). 

(2) Dip a borax bead into a solution of a manganous salt; fuse it in 
the oxidizing flame of the Bunsen burner, and cool (E 6 ). Repeat the 
fusion in the reducing flame of the Bunsen burner (E 6 ). 

(c) Manganous Acid. Manganese Dioxide. 

(1) Divide the mixture saved from (a) into two parts. Shake one part 
with air (E 7 ); to the other add a little 3% hydrogen peroxide solution, 
and warm gently. Filter, and test the solubility of the residue in cold, 
dilute sulphuric acid. Then add concentrated sulphuric or phosphoric 
acid, and warm gently (E 8 ) (R 740). 

(d) Manganates and Permanganates. 

(1) Fuse together some dry sodium carbonate, saltpeter, and a small 
quantity of a, manganous salt, on a piece of nickel or platinum foil (E 9 ). 

91 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

When cool, dissolve a portion of the product in a few cubic centimeters 
of water containing a drop of alkali. Dissolve the remainder in water, 
and add a drop of acid, or better, pass in carbon dioxide (Ew). What 
acid of chlorine behaves in a similar way ? 

(2) Place a pinch of lead dioxide in a dry test-tube. Add a little dilute 
nitric acid, and one drop of a solution of a manganous salt. Boil for a 
few moments, and allow the unchanged lead dioxide to settle (En). 

(3) Recall the actions of potassium permanganate on hydrogen sulphide, 
alcohol, and hydrogen peroxide (Ei 2 -Ei 4 ). 

(e) Burn tea, coffee, tobacco, etc., and boil the ashes with a little dilute 
nitric acid. Add a pinch of lead nitrate, and filter. 

Outside Questions. 

(1) Name elements which manganese resembles in each state of oxidation. 

(2) Why can manganic salts be prepared in concentrated sulphuric or phos- 
phoric acid solution, and not in dilute aqueous solution ? 

(3) Why does not hydrogen peroxide oxidize manganous salts as completely 
as does lead dioxide ? 



Experiment XLVIII. 
Manganese. 

LABORATOKY NOTES. 

Name Section 

Date ... Desk No. 



Experiment XLIX. 
Chromium. 

Discussion. Chromium exhibits three different valences in its various com- 
pounds. It is divalent in the chromous salts, trivalent in the chromic salts and 
the chromites, and hexavalent in the chromates. Its properties are very 
different in these three stages of oxidation. In the divalent condition it forms, 
for instance, a relatively strong base, and is of course a strong reducing agent; 
while in the hexavalent condition, on the other hand, it forms a strong acid, and 
is a strong oxidizing agent. 

It follows, therefore, that chromium will resemble various other elements, 
depending on its stage of oxidation. In the divalent condition it resembles 
divalent iron and tin. In the trivalent condition it resembles aluminum and tri- 
valent iron. In the hexavalent condition it resembles sulphur in the sulphates 
on the one hand, and molybdenum in the molybdates on the other. 

Directions. 

(a), (6) and (c). Repeat (d), (e) and (/) of Experiment XLIV, using chromic 
sulphate in place of aluminum sulphate. Answer similar question^ 
(EHE 4 ). 

(d) Add a pinch of sodium peroxide to 5 c. cm. of a solution of chromic sul- 
phate, and warm (E B ). Divide the solution among several test-tubes. 
To one add a solution of barium chloride (E 6 ). To a second add a solu- 
tion of lead nitrate (E 7 ). To a third add dilute nitric acid in excess (?) 
(E 8 ), and then add lead nitrate or barium chloride (E 9 ). 

(e) Recall the actions of hydrogen sulphide and sulphurous acid on a solu- 
tion of potassium dichromate containing sulphuric acid (Ei -E n ). 

Outside Questions. 

(1) To what acid of sulphur does dichromic acid correspond ? Why can 
potassium dichromate be considered an acid salt ? 



93 



Experiment XLIX. 

Chromium. 

/ 

LABORATORY NOTES. 

Name Section ... 

Date ... Desk No. 



Experiment L. 

Silver. 
Assigned Reading. Smith, pp. 626-635. 

Discussion. Silver is the noblest of the ordinary metals except for gold 
and platinum. At the same time, in certain properties, such as valence, and 
the crystalline structure of its salts, it resembles the alkali elements, the basest of 
the metals. These latter properties justify, so to speak, the position silver occu- 
pies in the periodic table. 

Directions. 

(a) Position of Silver in the Displacement Series. Place a strip of bright 
copper in a dilute solution of silver nitrate (Ei). What do you conclude 
as to the position of silver in the displacement series ? 

(6) Degree of Dissociation of Silver Hydroxide. Test a solution of silver 
nitrate for neutrality. To 5 c. cm. of this solution add a solution of 
sodium hydroxide, a little at a time, until no further precipitate is formed. 
Boil until the precipitate settles rapidly, then pour off the solution from 
the precipitate, and boil the latter repeatedly with fresh portions of water. 
After several such treatments, heat with a final portion of pure water, 
and test the resulting liquid with a drop of phenolphthalein solution. Col- 
lect the precipitate on a small filter paper. State two lines of argument, 
based on these experiments, regarding the strength of silver hydroxide 
as a base. 

(c) Silver Oxide. Transfer the precipitate obtained in (6) to a porcelain 
crucible, and heat over a flame, at first gently, but finally to the highest 
temperature you can attain with a Bunsen burner, or better, with a blast- 
lamp (E 2 ). Warm the residue with a few drops of concentrated nitric 
acid (E 3 ), and transfer to a bottle labeled " Silver Residues." 

(d) The Relative Solubilities of Some Silver Salts. 

(1) To a 2 c. cm. portion of silver nitrate solution add a solution of 
sodium chloride in slight excess. 

(2) To the resulting mixture add just enough ammonium hydroxide solu- 
tion to dissolve the precipitate (4). 

(3) To the resulting solution add about a cubic centimeter of a solution 
of sodium bromide, followed by 

(4) sufficient ammonium hydroxide solution to dissolve the precipitate. 

(5) To this solution add a cubic centimeter of sodium iodide solution, 
followed by 

95 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(6) a considerable quantity of ammonium hydroxide solution. If this 
does not dissolve the precipitate let it settle for a moment, and then 
pour off most of the solution and treat the precipitate directly with con- 
centrated ammonium hydroxide solution. 

(7) Pass a little hydrogen sulphide gas into the mixture, and finally 

(8) add an equal volume of potassium cyanide solution (E 5 ). 

(e) Preparation of Pure Silver. Dissolve a dime in a little warm nitric acid 
in the hood. Dilute the solution with water (?), add hydrochloric acid 
until no further precipitate is formed, then wash the precipitate several 
times with hot water, by decantation/and filter. Mix the moist precipi- 
tate with a little sodium carbonate and powdered carbon, transfer the 
mixture to a small cavity in a piece of charcoal, and heat it with the blast- 
lamp. When the action is over, if a bright globule is not obtained, transfer 
to a fresh cavity and heat again. 

1 See foot-note 1, page 109. 



96 



Experiment L. 
Silver. 

LABORATORY NOTES. 

Name Section ... 

Date Desk No. 



PART II. 

EXERCISES IN THE PREPARATION OF INORGANIC 

SUBSTANCES. 

General Instructions for Laboratory Work. 

The Instructions given at the beginning of Part I apply here also. The 
following additional matters require mention. 

Chemical Supplies. -- The materials for each preparation will usually be 
found on the end-desks. One important ingredient only, in each experiment, 
is to be obtained already weighed out at the store-room. The yields are to be 
calculated on the basis of this ingredient. 

Laboratory Notes. These are to be similar to those required for the 
Experiments in Part I. In addition, the notes must include data as to weights 
of materials used, the calculated, as well as the actual yields, the impurities found 
and their approximate amounts. A specially ruled sheet is inserted after each 
of the preparations on which entries may conveniently be made. 

For the sake of convenience, portions of the Procedure referred to in the 
Questions are occasionally indicated in parentheses by " Q," followed by an 
appropriate numeral. 

Laboratory Reports. Special Report cards are to be used for these 
Exercises, a separate card for each preparation. These will be furnished with 
your set of apparatus. Each preparation, properly labeled with one of the labels 
supplied for that purpose, and accompanied by the Report for that preparation, 
is to be handed to the instructor as soon as completed. 



97 



I. Potassium Nitrate from Sodium Nitrate and Potassium Chloride. 

Assigned Reading. Smith, p. 556. Thorp, pp. 142-147. 
Suggested Reading. Roscoe and Schorlemmer, Vol. II, p. 338. Eyde, 
Transactions, Congress of Applied Chemistry, Vol. 28, pp. 169-181. 

Discussion. The sodium nitrate from Chili and Peru, although threatened 
by the calcium nitrate made by the Birkland and Eyde " air burning " plants 
in Norway, is still, in general, the cheapest commercial source of nitrates. Potas- 
sium chloride, too, is the cheapest of soluble potassium salts. It is fortunate, 
therefore, that the solubilities in water of these two substances, and of sodium 
chloride and of potassium nitrate, are such that the potassium nitrate can be 
easily obtained by the proper evaporation of an aqueous solution of sodium 
nitrate and potassium chloride. 

The process can easily be understood if we consider that when the two salts, 
potassium chloride and sodium nitrate, dissolve in the water, they immediately 
dissociate very extensively into their ions, the more extensively the greater the 
dilution; that is, equilibria are at once established between the undissociated 
molecules of potassium chloride and of sodium nitrate, and their respective 
ions. However, since the ions when once formed are quite independent, some 
sodium and chlorine ions, and some potassium and nitrate ions must combine 
to form undissociated sodium chloride and potassium nitrate; that is, dissocia- 
tion equilibria between undissociated sodium chloride and potassium nitrate 
molecules and their respective ions are also established. These equilibria can 
be represented as follows: 

NaNO 3 2 Na+ + NO 3 ~ 

KCI 2 cr + K+ 

tl IT 

NaCl KN0 3 . 

Since all of the salts concerned dissociate to about the same extent in water, 
if the potassium chloride and sodium nitrate were originally present in the solu- 
tion in equal molecular amounts, that is, if the same number of gram molecules 
of each were originally dissolved in the water, then the undissociated molecules 
of all of these salts would be present in the mixed solution at about the same con- 
centration; and therefore, if such a solution were evaporated, that salt would 
first separate out which is the least soluble at the temperature of the evaporation. 
The solubilities of these salts at different temperatures are plotted as curves in 
the accompanying diagram (Fig. 4). 

99 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

It is evident that sodium chloride is the least soluble of these salts at the boiling 
point of water, and, therefore, if a solution such as that mentioned above is 
evaporated by boiling at atmospheric pressure, it will be the first to separate out. 
If this evaporation is continued until a nearly saturated solution of potassium 



20 




TEMPEBATURE 



100 



Fig. 4 



nitrate is obtained, and the solution is filtered while still hot, the clear filtrate as it 
cools must deposit large quantities of potassium nitrate, for the solubility of this 
substance falls off very rapidly with the temperature. 

The delay and expense of evaporation can be avoided by merely digesting 
the mixture of sodium nitrate and potassium chloride at 100 with just enough 
water to keep all of the resultant potassium nitrate in solution. This method is 
adopted in the following procedure. 

100 



I. POTASSIUM NITRATE 



Procedure. Fit up a 500 c. cm. Erlenmeyer flask so that it can conve- 
niently be handled when hot. This may be done by wrapping a strip of cloth 
around the neck of the flask and tying or wiring it on. Also prepare a filter for 
suction filtration. 1 

Heat 100 grams of sodium nitrate and 90 grams of potassium chloride with 
80 c. cm. of x water for ten minutes in the prepared Erlenmeyer flask. Keep the 
mixture just below the boiling point and shake frequently. While still hot, pour 
the liquid through a suction filter. [Caution. The filter flask, because of its 
thick walls and bottom, will not withstand sudden changes of temperature. 
Therefore warm it over the steam-bath before introducing the hot liquor, and 
do not allow the hot flask to rest directly on the cold sink.] Suck the residue dry, 
and while still hot transfer the filtrate to a small casserole. Warming the filter 
flask over the steam bath may help in this transference. Cool the casserole 
and contents to 10 by placing it in a larger vessel of cold water and stirring. 
Collect the resulting crystals on a fresh filter. Suck them dry, pressing them 
down with your spatula. Return the filtrate to the small casserole, and evap- 
orate until a considerable quantity of the sodium chloride has again separated 
out. Then filter this while still hot, and treat the filtrate as before. Combine 
both crops of potassium nitrate in a small beaker, and test for the presence of 
chloride by dissolving a few of the crystals in water, adding a little dilute 
nitric acid, and then a few drops of silver nitrate (Ei). A heavy precipitate 
of silver chloride will undoubtedly be obtained. The sodium chloride indicated 

1 Suction Filtration. When the precipitate is crystalline or granular, filtration may be carried out 
much more rapidly by the use of suction. For this purpose fit up a Hirsch funnel as in Figure 5. Place 
a disk of filter paper centrally on the filter plate, wet it with 
a jet of water from your wash-bottle, draw it firmly down on 
the plate by applying suction for a moment, and press the 
edges tightly against the sides of the funnel so that no free 
channel remains. Pour the liquid to be filtered on to the 
middle of the paper, directing it with a stirring-rod if neces- 
sary, so that it does not run down the sides of the funnel and 
turn back the edges of the filter paper. 

The suction is obtained with a so-called water pump, or 
aspirator. This pump, when operating properly, will lower 
the pressure in a closed vessel very nearly to the vapor pres- 
sure of water at the prevailing temperature, provided the water 
is supplied at a pressure somewhat greater than atmospheric 
(15 Ibs. per sq. in.). 

It is wholly unnecessary to keep the suction pump F I- 5 

working continuously; it wastes a great quantity of water, and makes noise. Apply the suction for two 
minutes, close off the filter flask with the pinchcock shown in the figure, and then turn off and disconnect 
the pump. Never, therefore, keep the pump going more than two minutes at a time. 

When only small quantities of material are to be filtered by suction, certain modifications should 
be made in this apparatus; the flask should be replaced by a test-tube with a side arm for attachment 
to the pump, and the Hirsch funnel should be replaced by an ordinary funnel provided with a small 
perforated cone of platinum or porcelain to support the tip of the filter paper folded in the ordinary 
conical fashion. 




101 



. 

LABORATORY MANUAL OF GENERAL CHEMISTRY 

by this test may be mostly or wholly removed by adding just enough cold 
water to moisten the crystals, stirring thoroughly so as to bring all parts in 
contact with the water, transferring to the suction filter, and then sucking dry, 
pressing with the spatula as before. (A little potassium nitrate is, of course^ 
sacrificed in this process. Do not, therefore, add more water than is necessary.) 
Again test a few of the crystals for the presence of chloride. If chloride is still 
present, recrystallize the crystals by dissolving them in one-half to three-quarters 
their weight of water, cooling, and filtering again on a fresh filter. Continue 
this treatment until the crystals do not give a test for chloride. Spread the 
crystals upon a clean porous plate, and allow to dry. Then weigh them, and 
place them in a clean specimen bottle, and label. 

Tests. --The dissolved salt should give no precipitate with silver nitrate. 
When held in the flame on a clean platinum wire, it should give the violet 
coloration characteristic of potassium, with none of the yellow sodium color. 
It should give no test for calcium (E 2 ), magnesium (E 3 ), or sulphate (E 4 ) ions. 

Outside Questions. 

(1) Name the ions and salts present in the solution of sodium nitrate and 
potassium chloride. 

(2) (A) Equal numbers of gram molecules of sodium nitrate and of potassium 
chloride are dissolved in an excess of water. What salt will separate 
out first when this solution is evaporated (1) at 100, (2) at 50, (3} at 
? (B) Suppose a solution containing twenty times as many gram 
molecules of potassium nitrate as of sodium chloride were evaporated at 
100, what salt would first separate out ? 

(3) What would you estimate to be the percentage amount of sodium chloride 
in your first precipitate of potassium nitrate, judging by the relative 
positions of the solubility curves in the diagram ? 

(4) On the basis of Fig. 4, devise a procedure for separating an equimolar 
mixture of potassium and sodium chlorides into the pure salts. 

(5) 100 grams of water dissolves the following amounts of sodium chlorate at 
different temperatures: 

20 ^ 40 80 100 

82 99 123.5 175 204 grams 

Draw the solubility curve for this substance in Fig. 4, and on this basis 
describe a procedure for obtaining potassium chlorate from sodium chlo- 
. rate. Reproduce these solubility curves in your Report. 



102 



I. Potassium Nitrate from Sodium Nitrate and Potassium Chloride. 



LABORATORY NOTES 



DATE. 



Materials 
used 


Grams 


Equations for Reactions Involved in the Preparation and Tests 


























Actual Yield 






Theoretical 
Yield 






Impurities 


Amount 




















% 


















Answers to Questions 



















































II. Pure Sodium Chloride from Rock Salt. 

Assigned Reading. Smith, pp. 570, 584, 585. 

Suggested Reading. Thorp, pp. 83-87. Roscoe and Schorlemmer Vol II 
p. 251. 

Discussion. Common rock salt usually contains the chlorides and sul- 
phates of potassium, calcium, magnesium, and iron as soluble impurities. Sodium 
chloride may be separated from most of these impurities by careful recrystalliza- 
tion, and a product suitable for table use may then be secured; But chemically 
pure sodium chloride is most easily obtained by passing hydrochloric acid gas 
into a nearly saturated solution of the impure salt. Sodium chloride is almost 
insoluble in concentrated hydrochloric acid, so that the greater part of it is 
precipitated as the hydrochloric acid gas is dissolved, and may subsequently be 
separated by nitration. The decreased solubility is, of course, partly due to the 
great decrease in the concentration of the sodium ion, resulting from the addition 
of chlorine ions; but molecular sodium chloride is also much less soluble in con- 
centrated hydrochloric acid than in water. 

Procedure. -- To 75 c. cm. of warm water add 25 grams of rock salt, and 
stir until the solution is saturated (Qi). The salt will dissolve more rapidly if it 
has been finely ground in a mortar. Add about one gram of sodium carbonate 
to the saturated salt solution. Stir, and allow the precipitate to settle (Q 2 ). 
Add a few drops of saturated sodium carbonate solution to the clear liquid; if a 
precipitate forms, it is evident that not enough carbonate was added the first 
time. When no further precipitate forms on the addition of a few drops of 
sodium carbonate solution, filter the salt solution through an ordinary plaited 
filter. 1 This will remove not only the precipitate, but any dirt or other insoluble 
matter contained in the rock salt. 

Now pass hydrochloric acid gas into the solution. For this purpose set 
up the apparatus shown in the accompanying figure (Fig. 6). It consists of a 
generator A supported on a ring stand, and a wash bottle B for purifying the 
hydrochloric acid gas as it is generated. The flask used for the generator should 
have a volume of at least one liter; a funnel tube D and a delivery tube E are 
fitted into its neck by means of a two-hole rubber stopper. The wash-bottle 
can be most simply constructed from an ordinary small filter flask. Into its neck 

1 A slimy, gelatinous, or very finely divided precipitate is best collected ivithoui suction. Suction 
forces such solid matter into the pores of the paper, so that liquid cannot pass through. For these 
precipitates it is most expeditious to use a plaited filter. Such a filter can be made very easily from an 
ordinary circular filter paper by proper folding along the radii of the circle. Ask your instructor to 
illustrate this method of folding if he has not already done so. 



103 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

fit the other end of the delivery tube, and a straight length of tubing F. Con- 
nect to its side tube a short-stemmed funnel G, dipping into the solution in the 
beaker C. Make sure that the apparatus is air tight. 




Fig. 6 

Place 50 grams of dry sodium chloride in the generator flask, and pour con- 
centrated hydrochloric acid into the wash-bottle to a depth of two centimeters. 
The tubes E and F must dip Below this acid. Use rubber tubing only when 
absolutely necessary; where it cannot be avoided, bring the ends of the glass 
tubes as closely together as possible. Place the saturated salt solution pre- 
104 



II. PURE SODIUM CHLORIDE FROM ROCK SALT 

pared above in the small beaker C, and adjust the funnel delivery tube so that it 
comes within one centimeter of the surface of the salt solution. Pour 75 c. cm. 
of sulphuric acid through the tube D. When the action appears to have ceased, 
warm gently. The frothing must not become so violent as to fill the generator 
and pass over into the wash bottle; careful heating will prevent this. Crystals 
of sodium chloride are formed as the gas bubbles through the salt solution. When 
no more sodium chloride is precipitated by the dissolving gas, remove the funnel 
delivery tube, and then stop generating the gas. 

Pour off the supernatant liquid, and remove the adhering mother liquor by 
suction filtration. 1 Dilute a small portion of the filtrate with several volumes of 
distilled water (Q 3 ), and test for sulphate. Discard the whole filtrate. If sulphate 
was found, wash the crystals on the filter by pouring a few cubic centimeters of 
concentrated, chemically pure hydrochloric acid over them. After each washing 
test the filtrate for sulphate as above. Continue washing until all the sulphate 
has been removed. Then suck the crystals dry, and transfer to a clean drying 
plate. Heat them gently on it until decrepitation ceases, then weigh, and place 
in a clean, dry specimen bottle. 

Tests. Dissolve a few of the crystals in distilled water; divide this solution 
into four test-tubes. To the first add ammonium oxalate (Ei) ; to the second, 
sodium phosphate and ammonium chloride (E 2 ); to the third, potassium fer- 
rocyanide (E 3 ); and to the fourth, barium chloride (E 4 ). i 

Outside Questions. 

(1) Why, although sodium chloride is but slightly more soluble in hot water 
than in cold, is it, nevertheless, advantageous to use hot water in dis- 
solving it ? 

(2) What impurities, if present, are removed by treatment with sodium car- 
bonate (E 5 ) ? What possible impurities would not be removed by this 
treatment ? 

(3) Why was a portion of the filtrate diluted with water before testing for 
sulphate ? 

(4) Explain the precipitation of the sodium chloride from its saturated solu- 
tion by means of hydrochloric acid from the point of view of the concen- 
tration law, writing equilibria equations. 

(5) Hydrocyanic acid is a gas, very soluble in water. Its dissociation con- 
stant in this solvent is 1.3 X 10~ 9 at 25. What would you expect to 

1 Concentrated hydrochloric acid attacks cellulose, the main constituent of filter paper, so that if 
an ordinary filter paper is used in this experiment, it will probably break when suction is applied, and 
will also contaminate crystals collected on it. A specially treated, so-called " hardened " filter paper 
may, however, be used successfully in this experiment. When hot hydrochloric acid is thus filtered, even 
this paper fails, and the Hirsch funnel is then best replaced by an ordinary funnel plugged with glass wool. 

105 



occur if this substance were bubbled through a nearly saturated solution 
of potassium cyanide ? Explain. 

(6) Assuming sodium chloride to be 50 % dissociated in its saturated solution, 
if hydrochloric acid had no effect on the solubility of the salt molecules 
as such, what would be the very greatest possible fraction of the total 
sodium chloride which could be precipitated by adding chlorine ions to the 
solution? 



106 



II. Pure Sodium Chloride from Rock Salt. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Actual Yield 



Theoretical 
Yield 



Impurities 



Grams 



Amount 



Equations for Reactions Involved in the Preparation and Tests 



Answers to Questions 



III. Strontium Chloride from Celestite (Strontium Sulphate). 

Assigned Reading. Smith, pp. 597-598, and 608-609. 
Suggested Reading. Roscoe and Schorlemmer, Vol. II, pp. 586. 589 
Thorp, pip, 427-428. 

Discussion. Celestite, SrSO 4 , is perhaps the most important source of 
strontium. It is a very sparingly soluble substance and is not easy to convert 
into other substances. One simple method for its treatment is to heat it with 
charcoal, whereby reduction to the sulphide is brought about. The sulphide, 
though difficultly soluble in water, dissolves readily in acids. 

Another method, which does not necessitate the use of high temperatures, 
depends on the insolubility of strontium carbonate. Finely ground celestite is 
converted into this substance by simply boiling with a strong solution of sodium 
carbonate. The carbonate, like the sulphide, as the salt of a weak and volatile 
acid, dissolves readily in most acids. 

This conversion of solid strontium sulphate into solid strontium carbonate 
by boiling with sodium carbonate is easily understood from the point of view of 
the concentration law. In a saturated solution of a binary electrolyte, such as 
strontium sulphate or carbonate, the equilibria 

[SrSOJ - SrS0 4 ? S0 4 + Sr++ 

solid solution 

exist, and according to the concentration law, 

c Sr+ + x C S Q 4 -- 



rSO4 



or <^sr + + X ^so 4 = Iv X 



Since the concentration of the strontium sulphate molecules in solution is kept 
constant, at constant temperature, by the crystals of the solid strontium sul- 
phate, the product of it and the equilibrium constant K, that is, K X C SrSO< , 
must also be constant. Hence, from the above equations, the product of the con- 
centrations of the ions, C Sr+ + X C S04 --, or the ion-product, as it is called, must 
also be constant. 

The numerical value of this ion-product can easily be determined for such 
a slightly soluble salt as strontium sulphate. Its saturated solution is so dilute 
that we may consider dissociation in it to be nearly complete, and so put the 
concentration of each ion as practically equal to the total concentration of the 
strontium sulphate expressed in gram molecules per liter. One liter of saturated 
strontium sulphate solution is found by experiment to contain 0.011 grams of the 
salt; therefore the total concentration is ^ = 0.00006 gram molecules per 

107 



liter. The strontium and sulphate ions evidently both have the same concen- 
tration, therefore, 

Car** X C S04 -- = (0.00006) X (0.00006) = 3.6 X 10~ 9 . 

Strontium carbonate is even less soluble; its saturated solution contains 
but 0.0011 grams per liter, or ^ or 0.000007 gram molecules per liter. Its 
ion-product is therefore, 

C S r" X C co ,-- = 4.9 X 10-". 

Expressed in words, these equations for the ion-product tell us that in the 
saturated solutions of strontium sulphate and of strontium carbonate the product 
of the concentrations of the ions must equal' 3.6 X 10~ 9 and 4.9 x 1Q- 11 respec- 
tively. If these values are exceeded, solid will crystallize out of the solution; 
if they are not reached, crystals will dissolve into the solution. 

The effect of the sodium carbonate can now be understood. Its addition 
in quantity to a saturated solution of strontium sulphate evidently gives an ion- 
product for strontium carbonate far in excess of its required value. Suppose, 
for instance, enough sodium carbonate were added to produce a normal solution 
of carbonate ions. The value of the ion-product, C Sr++ X C COl --, would then 
be 0.00006 X 1 = 6 X 10~ 5 , or over a million times greater than 4.9 X 10~ u . 
Strontium carbonate would therefore separate out of such a solution until the 
strontium ions were sufficiently reduced in concentration so that the ion-product 
had its required value. If solid strontium sulphate were present, it would imme- 
diately begin to dissolve, for with the removal of the strontium ions its ion- 
product, 3.6 X 10- 9 , is no longer reached. It would continue to dissolve, 
carrying, of course, sulphate ions into solution, until its characteristic ion-product 
(3.6 X 10- 9 ) had been reached. 

We can readily calculate what the relative concentrations of sulphate and 
carbonate must be in this final solution when both ion-products have been at- 
tained and thus a stationary condition reached. In this solution 

CST++ X C S04 -- = 3.6 X 10~ 9 ; and C Sr++ X C CO| -- = 4.9 X 10~. 
Therefore, dividing, 

Cso<- _ 3.6 X IP" 9 

Cco,-- ~ 4.9 X 10~ u * 

In any such solution then, where the sulphate ion is less than seventy-four 
times as concentrated as the carbonate ion, strontium sulphate will dissolve and 
strontium carbonate will precipitate; in any solution where it is more than 
seventy-four times as concentrated, strontium carbonate will dissolve and stron- 
tium sulphate will precipitate. Evidently then, heating strontium carbonate 
with a concentrated solution of sodium sulphate should convert it into sulphate. 
Experiment shows this to be the case. 



108 



III. STRONTIUM CHLORIDE FROM CELESTITE (STRONTIUM SULPHATE) 

Procedure. Grind 50 grams of celestite in a mortar until it no longer feels 
gritty under the pestle, and then transfer it to a 500 c. cm. casserole. Add to 
this 300 c. cm. of water and 65 grams of anhydrous sodium carbonate. Boil the 
mixture for at least 45 minutes, stirring at frequent intervals. Collect a small 
quantity of the solid residue on a filter, wash well with water, and test it for 
complete conversion into strontium carbonate. Continue boiling until the con- 
version is complete, then pour the mixture into a tall beaker, rinsing out the solid 
residue in the casserole with a little water. Allow the solid material to settle, 
pour off the supernatant liquid, and wash the residue three times with water by 
decantation. 1 To the washed residue add about 50 c. cm. of hot water and 
then add, a little at a time, 45 c. cm. of concentrated hydrochloric acid. Boil 
this solution for a few minutes, and then decant from any residue. If this 
residue contains strontium carbonate, boil it for a few moments with 1 c. cm. 
of concentrated hydrochloric acid, and combine the resulting solution with the 
main portion. Repeat this treatment until the carbonate is all removed from 
the residue. The one-tenth portion of strontium carbonate which was saved is 
now added. This will neutralize any excess of acid that may be present. Make 
sure that such is the case by testing with litmus after boiling for about five 
minutes. As the celestite used is liable to contain compounds of iron, a small 
portion of the solution should be removed at this point and tested for that 
substance. If iron is found, add a few drops of chlorine water, and boil for a 
few minutes (Qi). Filter the solution, and evaporate 2 until a slight scum forms 
when you blow across the surface of the liquid. Allow the solution to cool, stirring 
occasionally to prevent the formation of a solid cake. If the whole mass solidifies, 
it shows that the evaporation was carried too far. In that case, add a little 
water, redissolve the whole, and evaporate again, taking care not to evaporate 
as far as before. Drain the crystals formed on a suction filter. The mother 

1 When a precipitate is very insoluble and has a high enough specific gravity to settle rapidly in 
water, it can be most quickly and easily washed by decantation. That is, allow the precipitate to settle 
in a tall vessel, and pour (i. e., decant), or better still, syphon off the clear liquid. Fill the vessel again 
with fresh water, stir thoroughly and allow to settle, and again syphon off the clear liquid. The finer the 
precipitate, the longer the time required for settling. In this particular case (SrCO 3 ) the supernatant 
liquid is usually clear enough to decant after five minutes. 

2 Evaporation. The evaporation of aqueous solutions is best carried out upon a steam bath, since 
then no spattering occurs. The method is, however, very slow, and it is usually necessary to resort 
to more intense heating. For this purpose place the solution in a porcelain dish of suitable size, not 
in a glass beaker, for not only is glass too fragile, but the shape is not advantageous for a rapid removal 
of the vapor, and place the dish on a wire gauze over a flame. Take care that the flame is placed 
directly under the middle of the dish, so that it strikes only where the dish is in contact with the 
wire gauze. If the flame plays upon the exposed sides of the dish, above the level of the liquid, they 
will become very hot and may crack where they come in contact with the relatively cool liquid. Besides, 
the film of solution on the exposed sides will then evaporate rapidly and form a cake, and in some cases 
will be decomposed by the high temperature. Even at best, some solid crust may form around the edge 
of the liquid, and this should be scraped back into the liquid, or dissolved away by tilting. If spattering 
begins, remove the burner momentarily, and then continue heating with a smaller flame. Be especially 
watchful in this regard when the solution has become concentrated. 

109 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

liquor may be evaporated and a second crop of crystals obtained. Finally dry 
the crystals by spreading them on a porous drying plate and leaving them ex- 
posed for some time. Move them about frequently with a spatula, but do not 
heat (?). When they are so dry that they do not cling to one another, place 
in a specimen bottle and label. 

Tests. Prepare a roughly normal solution of your product. Using this solution, 

(1) test for neutrality with litmus; 

(2) test for iron with potassium sulphocyanate (Ei) ; 

(3) test for calcium by first precipitating the strontium with an excess of 
sodium sulphate solution (E 2 ), allowing the mixture to stand for twenty 
minutes, filtering, diluting with 5 volumes of water, and adding a little 
ammonium oxalate solution (E 3 ). 

Outside Questions. 

(1) Why was chlorine water added to the impure solution of strontium 
chloride (E 4 ) ? 

(2) Why was it important to have an excess of carbonate rather than an excess 
of acid ? 

(3) One liter of a solution containing one gram molecule each of sodium 
carbonate and sodium sulphate is heated for a long time with a large 
quantity of finely powdered strontium sulphate and strontium carbonate. 

(A) What will happen ? (B) Optional How much ? 

(4) Explain why in the test for calcium it was necessary to dilute the filtrate 
from the strontium sulphate before adding the ammonium oxalate. See 
the data on the inside of the front cover of Smith. 

(5) (A) From the data given on the inside of the front cover of Smith, calculate 

the ion-products of barium, strontium, and silver chromates. 

(B) What must be the ratio between the concentrations of the strontium 
and barium ions in a chromate solution such that neither strontium 
nor barium chromate will dissolve in it ? 

(C) What will happen if, say, 10 grams each of barium and strontium 
chromates are shaken with a liter of molar strontium nitrate solution ? 

(D) Optional. How much ? 



no 



III. Strontium Chloride from Celestite (Strontium Sulphate). 



LABORATORY NOTES 



DATE. 



Materials 
used 


Grams 


Equations for Reactions Involved in the Preparation and Tests 


























xVctual Yield 






Theoretical 
Yield 






Impurities 


Amount 






































Answers to Questions 



















































IV. Ammonium Iron Alum from Siderite. 

Assigned Reading. Roscoe and Schorlemmer, Vol. II, pp. 1149, 1215. 
Suggested Reading. Roscoe and Schorlemmer, Vol. II, p. 719. 

Discussion. A group of substances has long been known whose formulas 
can all be represented by the general formula: - 

M'M 111 (A VI 4 ) 2 12 H 2 

in which M 1 and M m represent mono- and trivalent metals, and A VI repre- 
sents the hexavalent acid elements sulphur or selenium. The commonest mem- 
ber of this group is KA1(S0 4 ) 2 12H 2 0, or ordinary alum, and has furnished a 
name for the whole group. The substance to be obtained in this preparation, 
for instance, has the formula (NH 4 )Fe(SO 4 ) 2 12H 2 0, and is known as ammonium 
iron alum. M 1 may be Li, Na, K, Rb, Cs, NH 4 , Tl, Ag, etc., while M m may 
be Al, Cr, Fe, Mn, Tl, etc. All alums have the same crystalline form, and will 
crystallize together in the same crystal, out of a solution saturated with respect 
to all of them. They are, in other words, isomorphous. A crystal of one alum 
dropped into a supersaturated solution of another alum, will serve as a nucleus on 
which the other alum will grow. By transferring such a crystal back and forth 
between evaporating solutions of differently colored alums, it is possible to build 
up beautifully variegated, banded crystals. 

Alums belong to the group of double salts, and can, in general, be made by 
evaporating solutions of the two component salts. The solubility curves for such 
a system, say of ferric sulphate, ammonium sulphate, ammonium iron alum, and 
water, can conveniently be represented by the following figure (Fig. 7). Dis- 
tances measured vertically upward in the figure represent the number of gram 
molecules of ferric sulphate in the solution per 1000 grams of water; distances 
measured to the right represent the number of gram molecules of ammonium 
sulphate in the solution per 1000 grams of water, all at constant temperature. 
The points A and F represent the solubilities of ferric sulphate and ammonium 
sulphate, respectively, in pure water. The curve AB represents the solubility of 
ferric sulphate in solutions containing different concentrations of ammonium 
sulphate; in other words, it shows the effect of ammonium sulphate on the solu- 
bility of ferric sulphate; it slopes inward because the solubility of ferric sulphate 
is, of course, decreased by the presence of a salt containing sulphate ions. The 
curve EF similarly represents the effect of ferric sulphate on the solubility of 
ammonium sulphate; it too slopes inward. The curve BCDE represents the 
solubility curve of the alum. Its general slope indicates what we should expect, 



111 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

namely, that in a concentrated solution of ferric sulphate only a little ammonium 
sulphate is required to cause precipitation of the alum; and similarly in a con- 
centrated solution of ammonium 
sulphate only a little ferric sul- 
phate is required to produce the 
same result. 

From the figure it is not 
difficult to see what must occur 
when a dilute, equimolecular so- 
lution of ferric sulphate and am- 
monium sulphate, represented 
by the point S in the figure, is 
evaporated. It will get more 
and more concentrated, follow- 
ing the dotted, 45 line OC in 
the figure, until the solubility 
curve of the alum is reached at 
C, and the alum separates out. 
This will continue as long as any 
solution remains. If, however, 
a solution containing two mole- 




Fig. 7 



cules of ammonium sulphate to one of ferric sulphate is evaporated, its course will 
be represented by the dotted line OD. Here again, alum will separate out when its 
solubility curve is reached at D, but, in this case, the relative composition of the 
solution will be altered as more and more alum separates out, so that the solution 
will get relatively richer and richer in ammonium sulphate, that is, the com- 
position of the solution will follow the curve DE until ammonium sulphate too 
begins to crystallize out at E. These two salts will continue to separate out until 
the solution is all evaporated. 

Procedure. Heat 40 grams of powdered siderite for several hours with 
125 c. cm. of dilute (3M) sulphuric acid and 50 c. cm. of water in a 500 c. cm. 
Erlenmeyer flask on the steam bath, or over a low flame. All but 3 or 4 grams of 
a mixed, black and white sandy material should dissolve (Ei). Avoid the loss of 
much water by evaporation. 1 When no more material will dissolve, heat more 
vigorously and evaporate the solution to a volume of 75 c. cm., and filter rapidly 
through a plaited filter, while still hot, into a small Erlenmeyer flask. Cool the 
filtrate in the flask by immersion in cold water or cracked ice for fifteen minutes. 

1 If too much acid is taken, or if too much water is allowed to evaporate, an amorphous anhydrous 
ferrous sulphate will separate out and interfere with the procedure. 

There is a very simple method by which one can tell with certainty wnen action on the carbonate 
has ceased. 



112 



IV. AMMONIUM IRON ALUM FROM SIDERITE 

Filter off the mother liquor by suction from the resulting crystals. Evaporate 
to half its original volume and return to the small flask. Cool and collect the 
crystals on the same filter with the first crop. Suck the whole dry and rinse 
several times with alcohol, sucking the crystals dry after each rinsing. Spread 
out the crystals on a watch glass, and when the alcohol has evaporated, weigh 
them. 

Dissolve the ferrous sulphate crystals by warming with one-fourth their 
weight of water to which has been added the calculated quantity of dilute (3M) 
sulphuric acid required for the conversion to ferric sulphate. 1 To this solution, 
when cooled to 50, add ten percent more than the calculated amount of con- 
centrated nitric acid 2 required to convert the ferrous sulphate into ferric sulphate 
(E 2 ). Heat to boiling, and test a minute drop of the liquid for ferrous ions by 
means of potassium ferricyanide (E 3 ). If ferrous ions are found, continue the 
addition of the nitric acid, a few drops at a time, until the oxidation is complete. 
To drive off the excess of nitric acid, evaporate the ferric sulphate solution until 
it becomes syrupy, and then add enough water to bring the volume up to that 
of the original solution of ferrous sulphate. Heat to boiling, and add the calcu- 
lated amount of ammonium sulphate required for the formation of the alum, 
dissolved in the same volume as that of the ferric sulphate solution. Cover the 
beaker containing the solution and put it aside until the next period. 3 Collect 
the crystals in a funnel, wash with a very little water, and allow to dry on a porous 
plate. Evaporate the mother liquor to one-third its bulk, and obtain another 
crop of alum. Combine this with the first lot, weigh, and place in a labeled bottle. 

Tests. Dissolve a few crystals of the alum in water and test for the 
various ions into which the constituent single salts dissociate (E 4 -E 8 ). Test 
also for calcium ions (Ee). 

Outside Questions. 

(1) What is ferrous sulphate called in commerce ? 

(2) What would be the formula of an alum made from chromium and potas- 
sium sulphates ? 

(3) Does ammonium iron alum dissociate in water in the same way as do the 
single salts of which it is composed ? 

V 

1 Ferrous sulphate precipitated at room temperature contains seven molecules of water of crystal- 
lization per molecule. 

2 The concentrated nitric acid is the so-called " constant boiling" acid with a specific gravity of 
1.42 at 25, and contains 68% of HNO 3 . 

3 Ammonium iron alum is a good illustration of a substance whose rate of crystallization is very 
slow. Several days are required for the excess of the alum to crystallize out of its supersaturated solu- 
tion. Ferrous sulphate, on the other hand, was deposited much more rapidly. . 

If a few large, well-formed crystals of alum are desired, cool the aboe supersaturated solution, 
and by means of a moistened thread attached to the cover glass, suspend^midway in it a small crystal of 
alum obtained from the instructor. 



113 



LABORATORY MANUAL OF GENERAL CHEMISTRY 



(4) Describe, and illustrate with a diagram similar to Fig. 7, what occurs as 
a solution containing three molecules of ferric sulphate to one of am- 
monium sulphate is evaporated at a constant temperature. 

(5) What would be the effect of using twice the weight of the ammonium sul- 
phate in the above preparation, everything else remaining the same ? 



114 



IV. Ammonium Iron Alum from Siderite. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretieal 
Yield 



Impurities 



Amount 



Answers to Questions 



V. Ammonio-Cupric Sulphate. 

Assigned Reading. Smith, p. 625. 

Suggested Reading. Roscoe and Schorlemmer, Vol. II, p. 435. 

Discussion. - - The preparation of ammonium iron alum in IV illustrated the 
formation of a double salt. It was found to behave in solution quite as would a 
mixed solution of the two component salts, that is, it was evidently dissociated 
in solution into ferric, ammonium, and sulphate ions. 

When ammonia is added to a solution of copper sulphate, or of any cupric 
salt, the basic copper salt which first forms redissolves on a further addition of 
ammonia. When this solution is treated with alcohol, a salt is precipitated of 
the formula CuCNHs^SC^B^O known as ammonio-cupric sulphate. This sub- 
stance, first described by Stisser in 1693 and advocated by him as a cure for 
epilepsy, illustrates a class of compounds very different from the alums. When 
it is dissolved in water, sulphate ions are formed, but, except in that particular, 
the solution behaves very differently from either a solution of ammonia or one 
of copper sulphate. The ammonia molecules are attached very firmly to the 
copper, so that when dissociation takes place the ion Cu(NH3) 4 ++ is formed. 
Such a compound ion, whose component parts do not readily dissociate, is called 
a complex ion. The properties of this complex ammonio-cupric ion are decidedly 
different from those of the ordinary cupric ion. Its color, for instance, is differ- 
ent, and very much more intense. 1 The termination -0 is given to ammonia to 
indicate this complex attachment. 

To obtain large, well-formed crystals of this salt, they must, of course, be 
formed slowly, and the alcohol must therefore be allowed to mix only by diffusion 
into the water. There must be no rapid mixing due to convection currents. 

Procedure. Pulverize 20 grams of crystalline copper sulphate, and dissolve 
in 75 c. cm. dilute ammonium hydroxide (sp. gr. 0.96; end-desk). If the salt 
does not dissolve to a clear solution, filter through an asbestos or glass-wool filter 2 
(Qi). Place 100 c. cm. of alcohol in a 250 c. cm. gas-collecting bottle. By means 
of a separatory funnel resting upon the bottom of the bottle, carefully run in 
25 c. cm. of water. This will form a layer below the alcohol. Do not run all of 
the water out of the funnel, but allow enough to remain to fill the stem of the 
funnel completely. Otherwise air bubbles may be forced into the alcohol-water 

1 The ordinary cupric ion may itself be a complex ion containing water molecules firmly attached 
to the copper atom, though we usually write it simply as Cu ++ . Indeed, it is generally agreed that most 
ions are hydrated, though whether but one definite hydrate predominates, say Cu(H2O)4 ++ or whether 
a whole series is formed, such as Cu(H 2 O) ++ , Cu(H 2 O) 2 ++ , Cu(H 2 O) 3 ++ , etc., is still not definitely known. 

2 See footnote 1, page 105. 

115 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

mixture and disturb the layers. Now pour the ammoniacal copper solution into 
the funnel, and allow it to run slowly on to the bottom of the bottle. This 
operation, too, must be done carefully so as not to mix the layers. Remove 
the funnel cautiously, and set the bottle away for a week. At the end of that 
time, long crystals will have been formed adhering to the sides of the gas bottle. 
Agitation of the liquid will mix the alcohol and water layers more completely 
with the formation of a fine crystal meal, but it is desirable to keep this separ- 
ate from the large crystals, so syphon the contents of the bottle carefully into 
a clean beaker. The large crystals will stick to the bottle unless you treat 
them too -roughly, and therefore may be separated in this way from the crystal 
meal. Scrape them into a small casserole, and, in order to remove the aqueous 
solution and the remainder of the crystal meal, wash with 10 or 15 c. cm. of 
alcohol to which 1 c. cm. of ammonia solution has been added. Rotate this 
mixture in the casserole and quickly pour off the liquid with the suspended, 
finely divided material, into a beaker. Repeat this treatment twice with 10 
c. cm. of alcohol, and then once with 10 c. cm. of ether. Spread the crystals 
on a filter paper, and when the ether is all evaporated, weigh them, and place 
them at once in a tightly stoppered specimen bottle. Stir the contents of the 
beaker and pour through a suction filter, pressing out the mother liquor thor- 
oughly with a spatula. Wash these crystals several times with the same liquids 
as were used with the larger crystals. Weigh, and stopper them at once in 
a separate specimen bottle. 

Tests. Compare the behavior of a solution of the product with that of a 
solution of copper sulphate towards, (1) NaOH, (2) BaCl 2 , (3) Fe, (4) H 2 S, 
(5) K 4 Fe(CN) 6 ; write all equations (Ei-E s ). Compare the smell of this solution 
with that of a solution of ammonia of corresponding strength. 

Outside Questions. 

(1) Why was it necessary to filter the ammoniacal copper solution through 
asbestos (R 623) ? 

(2) What ions are shown by the above tests to be present in a solution of 
ammonio-cupric sulphate ? 

(3) What do tests (2)-(5) indicate regarding the relative concentrations of 
cupric ions in solutions of copper sulphate and ammonio-cupric sulphate, 
and in saturated solutions of cupric hydroxide, sulphide, and f errocyanide ? 
What would be the effect of digesting freshly precipitated cupric hydrox- 
ide, sulphide, and ferrocyanide separately with a solution of ammonium 
hydroxide ? 

(4) Tabulate the differences between the ammonio-cupric ion and the cupric 
ion. 



116 



V. Ammonio-Cupric Sulphate. 



LABORATORY NOTES 



DATE 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



VI. Ammonium Bromide. 

Assigned Reading. Smith, pp. 265-269. 

Suggested Reading. Roscoe and Schorlemmer, Vol. II, pp. 372-376. 

Discussion. It would be simplest to make ammonium bromide by the 
neutralization of ammonium hydroxide with hydrobromic acid. It is possible, 
however, by means of a reaction first described by Balard, the discoverer of 
bromine, to make this substance directly from bromine, with equally good yields, 
- and bromine is much cheaper than hydrobromic acid. The cheaper method, 
of course, is preferred for commercial manufacture. 
When bromine is added to water, the equilibrium 

Br 2 + H 2 O ^ HBr + HBrO, 
or Br 2 + H 2 ^ H+ + Br~ + HBrO 

is immediately established, although the balance lies very much to the left, 
that is, but very little hydrobromic and hypobromous acids are formed. 1 If a 
base is present in, or is added to, the solution, then the hydroxyl ions remove 
the hydrogen ions very completely, and the balance is shifted far to the right; 
that is, most of the bromine is converted into bromine ions and hypobromite ions. 

Br 2 + H 2 O 
2NaOH 




With ordinary bases, when cold, this is all that happens, except that bro- 
mate is gradually formed ; with ammonium hydroxide, however, a further action 
occurs. In its solutions there is, in addition to the ammonium hydroxide and 

1 The concentration law applied to this equilibrium gives the equation, 

C H + X C Br - X C H BrO = K 

Cfin X CH 2 O 
and measurements have shown that K has a value of 9.4 X 10~ u at 25. 

From this it is easy to calculate the concentrations of hypobromous acid formed when bromine is 
dissolved in water, for it is evident that the equilibrium lies far towards the left, that is, only a little of 
the bromine is used up, and we can take CB^, the concentration of the free bromine at equilibrium, as- 
practically equal to the concentration of the total bromine. Suppose then that one-tenth of a gram 
molecule of bromine is dissolved in a liter of water. CBFJ = (very nearly) 0.1; CH,O = ^6, and 
from the above chemical equation, CH + = CBF" = CHBrO Therefore 
C* 

= K = 9.4 X ID' 11 , or C H BrO = 0.0008. 



0.1 

That is, the concentration of the hypobromous acid will be only about eight-tenths of one percent of the 
concentration of the free bromine. 

117 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

its ions, a considerable amount of dissolved ammonia, NH 3 , and this is a de- 
cided reducing agent, so it is at once attacked and oxidized by the BrO~, form- 
ing nitrogen gas, water, and bromine ions; 

3BrO~ + 2NH 3 - N 2 + 3Br~ + 3H 2 0. 

As the nitrogen gas escapes this action goes to completion. 
The whole action can then be represented as follows: 



3Br 2 + 3H 2 O 2 


6H+ 


+ 3Br~ + 


3BrO~ 


6NH 4 OH 2 


6OH~ 


+ 6NH 4 + 




T* 
6H 2 


2NH 4 OH 2 2H 2 + 

N 2 + 


2NH 3 


I 

3H 2 



3Br 



or, added together, 
3Br 2 



8NH 4 OH = 6NH 4 Br + N 2 +' 8H 2 0. 



It is seen that all the bromine atoms finally become bromine ions in solution, 
and, on evaporation, appear as ammonium bromide. 

Procedure. Take 100 c. cm. of aqueous ammonia (sp. gr., 0.94) or dilute 55 c. 
cm. of concentrated ammonia (sp. gr., 0.90) with 50 c. cm. of water in an Erlen- 
meyer flask, and cool by rotating the flask in a pail, or pan, of ice water. Examine 
the stopcock of the small separatory funnel, making sure that it is well lubricated 
with vaseline and that a rubber band holds it firmly in place, so that liquid in 
the funnel cannot leak on to your fingers. Attach the funnel to a retort stand 
with its stem extending into the Erlenmeyer flask, the flask in turn resting in its 
jacket of ice water. Pour 15.8 c. cm. of bromine into the small separatory funnel x 
and allow it to run, a few drops at a time 2 , into the ammonia. After each addition, 
rotate the flask in the ice water until the yellow color of the bromine has wholly 
disappeared. This procedure will not only prevent the addition of too much 
bromine, but what is more important, will prevent the solution from warming up 
and so making possible the formation of a dangerously explosive compound (QO . 
When the yellow color no longer disappears on shaking, stop the addition of 

1 Bromine produces very bad burns when it comes in contact with the skin. The best remedy is 
to wash off the bromine immediately under the tap, afterwards bathing the skin in very dilute ammonia. 
To avoid danger, have your instructor approve your apparatus before beginning the addition of the 
bromine. 

2 A convenient method of discharging small amounts of liquid from a dropping funnel is to give 
the stopcock a sudden turn through nearly 180. The number of drops delivered may be varied by 
varying the speed of turning. On your first trial, of course, make the turn a very rapid one. 

118 



VI. AMMONIUM BROMIDE 

bromine, and add a few drops of dilute ammonia until the solution is colorless. 
Evaporate the solution on the steam bath until only a little liquid remains, 
separate the crystals from the mother liquor by suction, and dry them on a 
porous plate, or between several large sheets of filter paper. 

Outside Questions. 

(1) What explosive compound could be formed in this experiment ? (Roscoe 

and Schorlemmer, Vol. I, pp. 516-519.) 
. (2) What fraction of the ammonia is sacrificed in this method of preparing 

the bromide ? 

(3) What concentration of hypobromous acid would be obtained by dissolving 
one-tenth of a gram molecule of bromine in a liter of water already con- 
taining hydrogen and bromine ions at unit concentration ? 



119 



VI. Ammonium Bromide. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



VII. Arsenic Acid. 

Assigned Reading. Smith, pp. 545-547. 

Suggested Reading. Roscoe and Schorlemmer, Vol. I. pp. 685-688- Vol 
II, pp. 221-222. 

Discussion. Arsenic forms two series of oxygen acids; one derived from 
the anhydride As 2 3 , the other from the anhydride As 2 O 5 . The most important 
representative of the first series is H 3 As0 3 , arsenious acid. This compound 
exhibits the properties which one would expect from the intermediate position 
of arsenic in the periodic table. It is a weak acid, its dissociation constant K 
at 25 being only 2xlO- 10 . It dissolves in strong acids, showing that it has some 
basic properties. It is thus an amphoteric substance. 

Ortho arsenic acid, or simply arsenic acid, H 3 As0 4 , is the chief representative 
of the second series. This acid, with a greater relative amount of oxygen in its 
molecule, is much more dissociated in solution than is arsenious acid. In other 
words, the more highly oxidized acid is the stronger; and this is usually the case 
among the different oxygen acids of a single element. 

Arsenic acid also has the properties which would be expected from its posi- 
tion in the periodic table. Arsenic acid and phosphoric acid are, for instance, 
remarkably alike. Both are very soluble in water and dissociate in it in the 
same way and to about the same extent, phosphoric acid being about 24% and 
arsenic acid about 19% dissociated in 1/24 molar solution. The anions of both 
are colorless, and form salts which exhibit similar solubilities and similar degrees 
of hydrolysis in water. Both acids form hydrates of the 'same formula, 
namely (H 3 PO 4 ) 2 H 2 and (H 3 As0 4 ) 2 H 2 O, which are isomorphous (Q 2 ). The 
former melts at 30, the latter at 35.5. Both acids when persistently heated 
lose water to form pyro and meta acids of similar formulas (E x and E 2 ). Arseni- 
ous acid is converted into arsenic acid by a variety of vigorous oxidizing agents. 
Iodine in aqueous solution, for instance, is often employed in analytical work 
for this purpose. 

H 3 As0 3 + I 2 + H 2 2 H 3 As0 4 + 2H^ + 2I~ 

This reaction is, however, reversible, so that if the arsenious acid is to be com- 
pletely converted into arsenic acid, the product of the concentrations of the other 
substances on the right hand side of the equation, that is C H + 2 X Cj- 2 must be 



121 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

kept small. This can readily be done by working in neutral or slightly alkaline 
solutions. 1 

In this preparation nitric acid is used as the oxidizing agent because it reacts 
energetically with the arsenious oxide, forms no stable compound with the result- 
ing arsenic oxide, and the products of its own reduction and decomposition are 
volatile. The solution of arsenic acid, secured by adding water to the arsenic oxide, 
is evaporated until it has the composition corresponding to the hydrate (H 3 AsO 4 ) 2 
H 2 0. The liquid thus obtained does not usually solidify when cooled below its 
melting point, exhibiting the phenomenon of supercooling in a pronounced man- 
ner, but " inoculation " of the melt with a tiny crystal of the hydrate, or of the 
corresponding isomorphous hydrate of phosphoric acid, will start a rapid solidi- 
fication, the latent heat of fusion being simultaneously given off, warming the 
system to its melting point. 

1 The equilibrium constant of this reaction has been determined, so that one can now easily com- 
pute the fraction of arsenious acid which can be oxidized in various mixtures of iodine and hydriodic 
acid. The equilibrium constant for this reaction, 

TT (CH.ASO,) (Ci,) (C H ,o) 

~ 77^ \~/ri s, /,-, ., = 1.1 at 25 . 

(CaAsoJ (Cr) 2 (C H + ) 2 

If there is no arsenic acid present at the start, the fraction of the arsenious acid which will escape 
oxidation will be approximately equal to the ratio of the concentrations of the two acids at equilibrium, 
that is, to 



But 

CH.ASO. = (1.1) (Cr) 2 (C H + ) 2 

C H ,As04 = (Cl 2 ) (C H20 ) ' 

or since CH,O = "F = 55.5 

CH.ASO. (.02) (Cr) 2 (C H + )' 



To illustrate the use of this formula, let us now suppose that we add .01 of a gram molecule of 
arsenious acid to a solution containing 1 gram molecule each of I 2 , I~ and H+. What fraction of the 
arsenious acid will escape oxidation ? According to the above, oxidation must proceed until 

CH.ASO. (.02) (I) 2 (I) 2 
- - , or 0.02, 



neglecting the small changes in the concentration of H + , I~ and I 2 which would necessarily occur. That 
is, about two parts in a hundred, or two percent of the arsenious acid would not be oxidized. 

Suppose, however, that the same amount of arsenious acid is added to a solution containing not 
only the iodide ion and the iodine molecule at the same concentrations as before, but also the salt of a 
weak acid in such quantity that the concentration of the hydrogen ion is kept nearly constant at one- 
thousandth normal. In this case 

CH.ASO, (.02) (1) (.001) 2 

rvT = 0.000,000,02. 

CHAs04 (I) 1 

That is, only about one part in fifty million of the arsenious acid would escape oxidation. 

This explains why arsenious acid can be completely converted into arsenic acid, so far as any 
chemical test will show, by oxidation with iodine in the presence of sodium hydrogen carbonate, where 
of course the hydrogen ion concentration is very small. 

122 



VII. ARSENIC ACID 

Procedure. To 50 grams of arsenious oxide in a large casserole add 25 
c. cm. of water; place the casserole under the hood, and add 75 c. cm. of concen- 
trated nitric acid. When the first action is over, warm gently from time to time 
just enough to keep the action going (E 3 ). When no more red vapors come off' 
heat strongly for two or three minutes. To determine the completeness of the 
oxidation, test for arsenious acid by diluting a few drops of the solution with a 
few cubic centimeters of water in a test-tube, adding an excess of sodium bicar- 
bonate crystals (E 4 ), and then a few drops of an iodine solution. If the color of 
the iodine disappears, arsenious acid is present and the oxidation is not complete 
(E 5 ). In this case, add more nitric acid and continue the warming. 

Finally evaporate the liquid just to dryness, rotating the casserole all the 
while until the residue is dry. Avoid heating the residue too strongly. Allow 
the casserole to cool, and then add 65 c. cm. of water. Dissolve the residue com- 
pletely by warming and stirring. Transfer the liquid to a smaller casserole, and 
evaporate until a thermometer dipped in it shows a temperature of 115. Trans- 
fer the liquid to a small beaker supported on a ring-stand in one of the sinks 
Evaporate further until a temperature of 160 is shown by the thermometer! 
Allow the liquid to cool to 60 and transfer to a dry, weighed, specimen bottle! 
Cool this bottle to 15 under the tap, taking care to get no water into the acid. 
Add to the liquid a small crystal of (H 3 As0 4 ) 2 H 2 0, obtained from the instructor 
in charge, and stir with the thermometer. Record the highest temperature 
attained, and observe how long this temperature is maintained. 

Tests. Dissolve about one gram of the arsenic acid hydrate in 20 c. cm. 
of water, and use this solution for the following tests: 

(1) Apply the test for arsenious acid outlined above. 

(2) Add a few drops of a dilute solution of potassium iodide (E). 

(3) Add a few drops of a solution of magnesium sulphate, and then add 
ammonium hydroxide until strongly alkaline (E 7 ). 

(4) Add a few drops of silver nitrate solution, and then ammonium hydrox- 
ide, a little at a time (E 8 and E 9 ). 

(5) Repeat (3) using phosphoric acid instead of arsenic acid (E 10 ). 

Outside Questions. 

(1) (A) What is meant by the term isomorphousf 

(B) What other isomorphous substances have been studied in Chemistry 

(C) Of what practical importance in chemical work is a knowledge of the 
isomorphic relations of substances ? 

(2) (A) Compare (E 6 ) with (E 4 ). Explain. 

(B) How would you provide the most favorable conditions for the 
complete reduction of arsenic acid to arsenious acid by means of 
hydriodic acid ? 

123 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(C) Optional. One-hundredth of a gram molecule of arsenic acid is 
added to a solution in which the hydrogen and iodide ions are at 
normal concentration, while the iodine is kept at a low concentra- 
tion, .0001 molar, by shaking the solution with a large quantity of 
a dilute solution of iodine in carbon disulphide. What fraction of 
the arsenic acid will be reduced to arsenious acid ? 



124 



VII. Arsenic Acid. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involyed in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



VIII. Cuprous Chloride from Copper. 

Assigned Reading. Roscoe and Schorlemmer, Vol. II, pp. 420-422. 

Discussion. Copper will not dissolve in either dilute or moderately con- 
centrated hydrochloric acid if air is excluded. It will dissolve in most acids if air 
or another oxidizing agent is present, or if the acid is itself an oxidizing agent as 
in the case of nitric and sulphuric acids. It will also dissolve slowly in concen- 
trated hydrochloric acid, even in the absence of an oxidizing agent. 

These facts are not difficult to understand when we recall the position that 
copper occupies in the displacement series of the metals. Copper will not dis- 
place hydrogen from a solution of its ions of normal concentration, but will, for 
example, displace mercury from a solution of its ions of similar concentration. 
Its position, therefore, is below hydrogen and above mercury. In the equilibrium 
then, between copper atoms and hydrogen ions, 

Cu + 2H+ Cu++ + H 2 , 

the balance lies far to the left. Only a minute quantity of copper ions and of 
hydrogen molecules is formed, not nearly enough to produce a perceptible 
pressure of hydrogen gas. If, however, air or some other oxidizing agent is pres- 
ent, even this minute quantity of hydrogen will combine with the oxygen to 
form water. This will disturb the balance of the equilibrium, more copper will 
dissolve to make good the loss of hydrogen, and this process will be repeated 
until either the copper, the acid, or the oxidizing agent is all consumed, or, more 
accurately, has become so dilute that it will no longer react. 

It is also easy to see from the above equilibrium equation why copper dis- 
solves in concentrated hydrochloric acid even in the absence of oxidizing agents. 
It is known that in concentrated solutions of chlorides the cupric ions form very 
stable complex ions with the chlorine ion; 

Cu+++3Cr^(CuCl 3 )". 

These chlorine ions remove the cupric ions from the above equilibrium just as the 
oxidizing agent removed the hydrogen gas, and the equilibrium is shifted in the 
same way; copper dissolves to make good the loss of cupric ions, the pressure of 
the hydrogen gas soon becomes equal to that of the atmosphere, and we observe 
its escape from the solution as bubbles. This action will continue until the acid or 
the copper is all consumed, or until the complex ions have become so concentrated 
that no more cupric ions can be removed. 

Whenever copper does dissolve in the presence of an oxidizing agent, even 
when the copper is in excess, it forms divalent, cupric ions, rather than monova- 

125 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

lent, cuprous ions. This unexpected behavior is due to the peculiar fact that it 
is easier to oxidize the cuprous ion to the cupric ion than to oxidize the copper 
atom to the cuprous ion. In other words, the cuprous ion, although already 
loaded with one charge of positive electricity, has a greater affinity for an addi- 
tional charge than has the original uncharged copper atom for the first charge. 
This relation is exactly the reverse of that between the ferrous and the ferric ions. 
As a consequence of this peculiar fact, cuprous salts are unstable in solution. 
One cuprous ion captures the electric charge from one of its fellows, so that 
metallic copper and cupric ions result ; 

Cu+ + Cu+ 2 Cu + Cu++. 

Cuprous ions can therefore exist only in very dilute solutions, or in solutions 
containing a high concentration of cupric ions. No easily soluble simple cuprous 
salts have been prepared. 

In this preparation, the balance of the equilibrium between cuprous ions, 
copper atoms, and cupric ions, as above written, is made to shift to the left by the 
presence of concentrated hydrochloric acid; the complex (CuCl 2 )~ which the 
chloride ions form with cuprous ions is even less dissociated than the corre- 
sponding complex (CuCl 3 )~ formed with cupric ions. The cuprous ions are 
thereby very largely eliminated, and the reaction proceeds to the left until either 
the supply of copper atoms, or of cupric ions is exhausted (the latter is the case 
here), or until the complex ions attain a concentration such that no further action 
occurs. 

Procedure. Add 13 grams of copper gauze, chips, or nails to 75 c. cm. of 
concentrated hydrochloric acid (sp.gr., 1.2) mixed with 75 c. cm. of water and 10 c. 
cm. of concentrated nitric acid in a 500 c. cm. flask. Observe whether any action 
occurs, then add a fragment of sodium nitrite the size of a pea, and place the flask 
under the hood (Ei) (Qi and Q 2 ). If all the copper does not ultimately dis- 
solve in this mixture, add a second 5 c.cm. portion of nitric acid. When the cop- 
per is wholly dissolved, pour the solution into a casserole, rinse the flask out with 
a little water and add this to the main solution. Evaporate nearly to dryness 
under the hood (best on the hot-plate, especially toward the end of the evapora- 
tion), add 50 c. cm, of concentrated hydrochloric acid and an equal volume of 
water, and evaporate again until crystals appear (E 2 ). 

Transfer these crystals and their mother liquor to an Erlenmeyer flask. Add 
50 c. cm. of water and 75 c. cm. of concentrated hydrochloric acid, using these 
liquids to rinse out the casserole. Charge 20 grams of copper gauze into the 
Erlenmeyer flask, and insert a small flask upside down in its neck to prevent easy 
access of air. Digest the mixture on the hot-plate at a temperature just below 
the boiling point until the green color of the solution has disappeared. This 
condition is usually reached in about an hour, and indicates that the reduction 
is complete. To make sure of this, remove a few drops of the solution by means 

126 



VIII. CUPROUS CHLORIDE FROM COPPER 

of a glass tube, and add a test-tube half-full of water. If the supernatant liquid 
left after the settling of the white precipititate is colored blue, the reduction is 
not complete. If it appears colorless, but on the addition of ammonia acquires a 
decided blue color, the reduction is nearly, but not quite, complete. 

When the reduction is complete, pour the liquid contents of the flask, while 
still hot, through a funnel loosely plugged with asbestos or glass wool, into a tall 
bottle or cylinder containing about a liter and a half of cold water. If a white 
residue remains in the flask, dissolve it in a little hydrochloric acid (1:1) and 
pour this through the funnel. 

Allow the white precipitate to settle, syphon off the supernatant liquid, and 
wash twice by decantation l with 200 c. cm. of 1 /2 molar sulphuric acid. Transfer 
the precipitate to the suction filter, and wash it there with several fresh portions 
of this same dilute acid. Finally wash it with two successive 20 c. cm. portions 
of alcohol, followed by two successive 10 c. cm. portions of ether. Do not, except 
when working with ether, suck the precipitate dry; as soon as the level of the 
washing fluid has reached the surface of the cuprous chloride, add the next quan- 
tity of washing fluid. 2 When the last ether washing has been sucked dry, transfer 
the cuprous chloride at once to a watch glass on the hot-plate, pulverize, and as 
soon as the smell of ether has disappeared, bottle, weigh, and label. 

Tests. 

(1) The product should be white. A small sample placed in a test-tube 
and covered with dilute ammonia should dissolve to a clear and, at first, 
nearly colorless solution. Shaken with air it should turn blue (E 3 ). 

(2) Dissolve a pinch of dry cuprous chloride in a very little warm, concen- 
trated hydrochloric acid (E 4 ). Add an excess of concentrated sodium 
hydroxide (E 5 ). Boil (E e ). 

Outside Questions. 

(1) How many cubic centimeters of concentrated nitric acid would be re- 
quired to oxidize the 13 grams of metallic copper to cuprous chloride, in 
the presence of the hydrochloric acid, if the nitric acid had a sp.gr. of 1.4 
and contained 65% of HN0 3 , assuming nitric oxide to be the sole 
reduction product from the nitric acid ? 

(2) Explain the action of the sodium nitrite on the rate of solution of the 
copper. 

^See footnote 1, page 109. 

2 These washing operations are the critical steps in the procedure, because air in contact with 
moist cuprous chloride oxidizes it. Carry out the washing therefore as rapidly as is consistent with 
thoroughness. 

The danger of oxidation may be greatly lessened if the water into which the cuprous chloride 
solution is poured and the dilute sulphuric acid used for washing are freshly boiled and then cooled. 
Similarly if a rapid stream of carbon dioxide is fed into the mouth of the funnel during filtration, it will 
form an excellent protecting blanket. For this purpose suitable carbon dioxide generators, ready 
for use, may be secured on temporary order from the store-room. 

127 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(3) Write out the equilibria equations showing the action of metallic copper 
on cupric chloride in concentrated hydrochloric acid. 

(4) If the complex (CuCl 3 )~ which hydrochloric acid forms with cupric ions 
were kss dissociated than the complex (CuCl 2 )~ which it forms with 
cuprous ions, what would be the effect of heating a solution of cuprous 
chloride in concentrated hydrochloric acid ? 



128 



VIII. Cuprous Chloride from Copper. 



LABORATORY NOTES 



DATE 









Materials 
used 


Grams 


Equations for Reactions Involved in the Preparation and Tests 


























Actual Yield 






Theoretical 
- Yield 






Impurities 


Amount 
































>. 






Answers to Questions 



















































IX. Mercuric Chloride, HgCl 2 (Corrosive Sublimate). 

Assigned Reading. Smith, pp. 655-656. 

Suggested Reading. Roscoe and Schorlemmer, Vol. II, pp. 674-681. 

Discussion. Mercury resembles copper in forming two series of salts, the 
monovalent and the divalent. The relation between the two series is, however, 
different for the two metals. A copper atom takes up a second charge of elec- 
tricity more easily than the first, so that energy can be gained when a singly 
charged copper atom transfers its charge to another singly charged atom; there- 
fore cuprous salts spontaneously decompose into cupric salts and metallic 
copper, except in very dilute solution. The mercury atom behaves in the 
opposite, and indeed more usual, way; it is more difficult to add a second charge 
of electricity than it was to add the first. Mercurous ions are therefore stable 
in solution. They require an oxidizing or reducing agent to convert them into 
the mercuric state or into metallic mercury. 

Cuprous and cupric ions both form complex ions with great ease, the cuprous 
ion being the more active in this respect. With mercury the reverse relation 
again holds; mercurous ions show a slight, mercuric ions a pronounced, tendency 
toward the formation of complexes. 

Mercury as a noble metal, of course requires an oxidizing agent to bring it 
into solution in acids. In this preparation it is easy to decide when oxidation 
is complete, for chlorine ions are present in solution, therefore if any mercurous 
ions are also present, a white precipitate of mercurous chloride will be formed. 

The mercuric chloride is purified by taking advantage of its volatility. It 
melts at 283, and boils, under atmospheric pressure, at 303, so that it is easy 
to distill it from many impurities. Moreover, the solid has a high vapor 
pressure even before it melts, so that at a temperature not too far below the 
melting point a slow sublimation will take place. This illustrates an important 
method of industrial purification. 

Procedure. Place 25 grams of mercury in a 250 c. cm. flask, and add 
30 c. cm. dilute nitric acid, and 20 c. cm. concentrated hydrochloric acid. Heat 
gently over a Bunsen flame (hood) until the action ceases (Ej. If a residue of 
mercury remains, add a little more of both acids; if a residue of calomel remains, 
add a little more nitric acid, and continue the heating. When a clear solution 
is obtained, transfer to a small casserole and evaporate to dryness. Clean and 
dry the outside of a 500 c. cm. flask, fill it partly with cold water, and place it 
on top of the casserole. Heat the casserole carefully so that a slow sublimation 
takes place. The sublimed mercuric chloride will adhere to the bottom of the 

129 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

flask in long, snow-white crystals; " white, dense and ponderous," the alchemist 
Geber called them. When the sublimation is complete, remove the flask from 
the casserole, scrape the crystals on to a paper and transfer them to a dry, 
weighed, stoppered, and labeled bottle. 

Tests. 

(1) The crystals should be soluble in hot water, and in cold, dilute alcohol, 
without residue. 

To a dilute solution of the crystals : - 

(2) Add a few drops of sodium hydroxide solution (E 2 ) ; 

(3) Add an equal volume of potassium chloride solution, and then a solution 
of potassium iodide, a few drops at a time, until no further change occurs 
(Es) ; show by a test that mercuric sulphide gives an even smaller con- 
centration of mercuric ions than does the complex ion here formed (E 4 ) ; 

(4) Add a drop of stannous chloride solution (E 6 ) ; 

(5) Add a clean copper nail (E 6 ) ; determine whether any copper goes into 
solution. 

Outside Questions. 

(1) (A) From the statements in the Discussion would you expect calomel 

or corrosive sublimate to experience the greater relative increase in 
solubility when sodium chloride is added to an aqueous suspension 
of the two salts ? 

(B) What property of mercuric ions does Test (3) illustrate ? What 
do you infer from this test regarding the relative stability of the 
chloride and iodide complexes formed ? Is this a usual distinction 
between chlorides and iodides ? (See Smith, pp. 237 and 651, for 
instance.) 

(2) (A) What is the action of an aqueous solution of mercuric chloride on 

litmus ? 

(B) What does this indicate regarding mercuric hydroxide ? 
((7) How does mercuric hydroxide compare with mercurous hydroxide 

in this respect ? 



130 



IX. Mercuric Chloride, HgCl 2 (Corrosive Sublimate). 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



X. Hydrazine Sulphate. 

Assigned Reading. Roscoe and Schorlemmer, Vol. I, pp. 506-509. 

Discussion. --This important substance was formerly prepared from am- 
monium thiocyanate, NH 4 SCN, by a difficult and roundabout method, and was 
correspondingly high-priced. It is now manufactured by a method discovered 
by Raschig in 1907, which is not only much simpler, but also uses much cheaper 
materials; so that the cost of hydrazine sulphate has been considerably reduced. 

Raschig's method produces hydrazine by the action of sodium hypochlorite 
on ammonia. This may appear surprising, for when sodium hypochlorite and 
ammonia are brought together in concentrated solution under ordinary condi- 
tions, a rapid evolution of nitrogen occurs, with the simultaneous formation 
of sodium chloride and water, without a trace of hydrazine, according to the 
equation, 

2NH 3 + 3Na+ + 3 OOP = N 2 + 3Na+ + 3OT + 3H 2 0. 

In dilute solution, however, using a gram molecule of hypochlorite per gram 
molecule of ammonia, this evolution of nitrogen does not occur, at least not for 
some time. The smell of ammonia nevertheless disappears, and the solution 
will not give a purple color when treated with aniline, a characteristic test for 
hypochlorites. Investigation shows that monochloramine, NH 2 C1, is formed in 
this reaction, and by careful distillation under low pressure it can be obtained 
as a pale-yellow, oily liquid, with a frightfully irritating odor. The equation 
for the reaction is 

NH 3 + Na+ + OOP = NH 2 C1 + Na+ + OH~, 
one equivalent of sodium hydroxide being formed. 

Monochloramine reacts further in two ways. It is attacked by ammonia 
to form nitrogen, and ammonium chloride; 

3NH 2 C1 + 2NH 3 = N 2 + 3NH 4 C1. 

This explains how the gaseous nitrogen is at once formed in the concentrated 
solutions. Ammonia also adds directly on to the monochloramine, 

NH 2 C1 + NH 3 = NH 2 NH 3 C1, 

forming hydrazine chloride. This second reaction is the useful one for our 
purpose, and the success of the preparation depends on hastening it at the 
expense of the first reaction. This is accomplished in two ways; first, by in- 
creasing the concentration of the ammonia, which accelerates the second reaction 
more than the first; second, by increasing the viscosity of the solution by the 

131 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

addition of such substances as glycerine, starch, white of egg, glue, gelatine, etc., 
which appear to hinder greatly the evolution of gas. 1 By the combined use of 
these two effects, a 70% yield of hydrazine chloride can be obtained. The 
chloride is easily soluble in water, but on evaporating its solution, cooling, and 
adding sulphuric acid, the less soluble sulphate precipitates. 

Procedure. 

(a) Preparation of a Molar Solution of Sodium Hypochlorite. Set up, 
under the hood, as a chlorine generator, a 250 c. cm. flask, into the neck of 
which a funnel-tube and a delivery tube bent twice at right angles have been 
fitted. The arrangement should be similar to that in Fig. 6, except that the 
wash-bottle and short-stemmed funnel are omitted. It will also be advisable 
to use a retort-stand instead of a tripod to support the flask, utilizing a second 
ring surrounding the funnel-tube to prevent the generator flask from tipping 
over. Put 13 grams of potassium permanganate in the generator, and add 50 
c. cm. of water. Warm until the salt is dissolved. 

Arrange for the absorption of the chlorine as follows. Dissolve 17 grams of 
sodium hydroxide in 35 c. cm. of water in a 500 c. cm. flask. Cool and add 125 
grams of cracked ice. Wrap a towel around the flask, and then insert the delivery 
tube from the chlorine generator into the flask, making sure that the tube is long 
enough to reach nearly to the bottom. Measure out 90 c. cm. of concentrated 
hydrochloric acid in a small flask, and pour it, a little at a time, into the funnel 
tube of the generator (Ei) . This must be done so slowly that no chlorine escapes 
absorption by the caustic soda solution (E 2 ). Test for escaping chlorine every 
little while by smelling guardedly at the mouth of the flask. Rotate the ab- 
sorption flask frequently, so that the caustic alkali will swirl up on to the walls 
of the flask and thus promote absorption. When the acid is all added and 
action has almost ceased, heat the generator gently (but not hotter than the 
hand can bear) and shake. This should complete the action and liberate most 
of the chlorine dissolved in the liquid. Remove the absorption flask, and add 
50 c. cm. of water. This should give an approximately normal solution of sodium 
hypochlorite. 

(6) Preparation of Monochloramine in Solution. Add 40 c. cm. of this 
sodium hypochlorite solution prepared in (a) to 100 c. cm. of a 1/2 molar solu- 
tion of ammonia (E 3 ). . Observe the odor. 

1 That these substances do slow down the decomposition reaction by increasing the viscosity is 
also indicated by the fact that acetone and similar substances which lessen the viscosity of the solution 
hasten the decomposition reaction to such an extent that scarcely any hydrazine chloride is formed. 

Another explanation of the activity of these substances has also been given, namely, that the starch, 
gelatine, etc., destroy or render inactive some catalytic impurity, originally present, which normally 
accelerates the first reaction. This explanation is supported by the fact that traces of metallic salts 
added to the solution hasten the decomposition, and that glycerine, albumen, etc., form complex com- 
pounds with these metallic salts. 



X. HYDRAZINE SULPHATE 
Tests. 

(1) Add two drops of the sodium hypochlorite solution to a few cubic centi- 
meters of water in a test-tube, and to this solution add a few drops of 
aniline water. 

(2) Repeat Test (1) using a few drops of the solution of monochloramine in 
place of the sodium hypochlorite solution. 

(3) Dilute 10 c. cm. of concentrated ammonia with an equal volume of water, 
add 20 c. cm. of the hypochlorite solution, and divide the resulting solu- 
tion among four test-tubes. Keep one portion for a blank; to a second 
(a) add a few drops of copper sulphate solution; to a third (6) add a few 
drops of acetone; and to a fourth (c) add a few drops of 1% gelatine 
solution. 

(c) Preparation of Hydrazine Sulphate. Dilute 125 c. cm. of concentrated 
(sp.gr., 0.90) ammonia, with 75 c. cm. of water in a 500 c. cm. Erlenmeyer flask. 
Dissolve Yi c, cm. of liquid glue in 5 c. cm. of water, and add this to the 
dilute ammonia solution. Mix thoroughly, then add 100 c. cm. of the freshly 
prepared sodium hypochlorite solution. At once heat this mixture rapidly to 
boiling under the hood (E 4 ). Be ready to remove the burner instantly if the 
liquid threatens to foam out of the flask. Continue the boiling for half an hour, 
when the solution should have evaporated to about one-third of its original 
bulk, and the odor of ammonia should have disappeared. If the solution is 
not perfectly clear, filter while hot through a plaited filter. 1 Dilute to a volume 
of 115 c. cm., cool in a flask under the tap (caution!}, add 10 c. cm. of con- 
centrated sulphuric acid (E 5 ), and pack in ice. After fifteen minutes, filter 
with suction, using a hardened filter in an ordinary funnel. 2 Dissolve these 
crystals in as little boiling water as possible, cool under the tap (caution!), add 
5 c. cm. of concentrated sulphuric acid, and cool in ice. Collect the crystals 
as before, wash with a little alcohol, followed by ether. Spread out the filter 
paper, and when the ether has fully evaporated, transfer the crystals to a weighed 
and labeled bottle. 

Tests. 

(4) Dissolve one-fourth gram of hydrazine sulphate in 25 c. cm. of water. 
(a) Add a few drops of this solution to a little very dilute ammouiacal 
copper sulphate solution and warm (E 6 ) . 

(6) Add a few drops of solution to a few cubic centimeters of 0.1 molar 
silver nitrate solution and boil (E 7 ). 

1 See footnote 1, p. 103. 

2 For working with small quantities of material this arrangement is preferable to a large Hirsch 
funnel. The hardened filter paper will withstand the heavy suction. Ordinary filter paper may be 
substituted for the hardened paper if a small perforated platinum cone is first placed in the funnel to 
act as a support. See footnote 1, p. 101. 

133 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(c) Add 6 molar ammonium hydroxide solution carefully, a drop at a 
time, to- 10 c. cm. of 0.1 molar silver nitrate solution until the silver 
hydroxide at first precipitated just dissolves (E 8 ). Then add one- 
half cubic centimeter more 6 molar ammonium hydroxide solution. 
Mix thoroughly, and add one cubic centimeter of the hydrazine 
sulphate solution (E 9 ). 

Outside Questions. 

(1) What are the relative strengths of hydrazine hydrate (hydrazonium 

hydroxide) and ammonium hydroxide as bases ? 

(2) What are the relative strengths of ammonia and hydrazine as reducing 
agents? Cite several instances to prove this. 

(3) Calculate how great an excess of sodium hydroxide was present in the 
sodium hypochlorite solution, assuming that the potassium permanga- 
nate was completely reduced. 

(4) What were the relative numbers of gram molecules of ammonia and of 
sodium hypochlorite used in the preparation of monochloramine? 

(5) What do this answer to Question (4) and the results of Tests (1) and 
(2) prove regarding the formula of monochloramine? 



134 



X. Hydrazine Sulphate. 



LABORATORY NOTES 



DATE. 



used 


Grams 


Equations for Reactions Involved in the Preparation and Tests 


























Actual Yield 






Theoretical 
Yield 






Impurities 


Amount 
































* 






Answers to Questions 



















































XI. Mercuric Thiocyanate. 

Assigned Reading. Smith, pp. 508 and 757. 

Discussion. --Thiocyanic acid, HSCN, or sulphocyanic acid, as it is 
sometimes called, is a volatile, unstable liquid at ordinary temperatures. It 
mixes with water in all proportions, forming colorless solutions which are rela- 
tively stable, and which are good conductors of electricity. Evidently then, 
HSCN is a strong acid, and its anion SCN~ is colorless. Its silver, mercurous,' 
and cuprous salts are insoluble, while its lead, mercuric, and cupric salts are but 
slightly soluble in water. In color, strength, and in the solubilities of its salts this 
acid evidently closely resembles the halogen acids. 

Since mercuric thiocyanate is so slightly soluble in water, whenever a solu- 
tion of a soluble thiocyanate is added to a solution of a soluble mercuric salt, a 
precipitate is, of course, to be expected; 

2KSCN + Hg(N0 3 ) 2 ^Hg(SCN) 2 + 2KN0 3 . 

A precipitate does usually form, but the precipitation is not complete except 
when equivalent quantities of the two salts are used. During the addition of 
thiocyanate to the mercuric salt, more or less of the precipitate is held in solu- 
tion by the mercuric nitrate, and the precipitation is not complete until this 
has all been used up. On the other hand, if an excess of thiocyanate is added, 
the precipitate partially or wholly redissolves. Mercuric thiocyanate is evi- 
dently soluble in an excess of either mercuric or thiocyanate ions. This is, of 
course, just the reverse of what would be expected from the concentration law, 
and resembles, except that it is even more pronounced, the solubility of cuprous 
chloride in hydrochloric acid. 

Investigation has shown that the explanation of both of these anomalous 
solubilities is the same, namely, the formation of complex ions. With an excess 
of mercuric nitrate a complex cation [Hg 2 (SCN) 2 ] ++ is formed; 

Hg(SCN) 2 + Hg++ + 2N(V [Hg 2 (SCN) 2 ]++ + 2NO 3 ~. 

With an excess of potassium thiocyanate a complex anion [Hg(SCN) 4 ]~ ~ is 
formed; 

Hg(SCN) 2 + 2K+ + 2SCN- [Hg(SCN) 4 ]- ' + 2K + . 

As these complex ions are very slightly dissociated into the simpler ions, and 
the compounds [Hg 2 (SCN) 2 ] (NO 3 ) 2 and K 2 [Hg(SCN) 4 ] are both soluble, the 
precipitate will evidently dissolve in the presence of either mercuric nitrate or 
potassium thiocyanate. 

135 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

In the preparation of mercuric thiocyanate, it is evidently essential that 
exactly equivalent amounts of the mercuric nitrate and potassium thiocyanate 
should be taken. One simple method to accomplish this involves the use of 
ferric chloride. A few drops of a solution of this substance are added to the 
mercuric nitrate solution; as potassium thiocyanate is now added, the solution 
remains colorless until the Hg ++ ions are all precipitated, after which SCN~ 
ions begin to combine with the Fe+ ++ ions to form ferric thiocyanate, an in- 
tensely red substance. The first appearance, then, of a red color is a signal to 
stop the addition. 

Procedure. Heat 25 grams of mercury in a flask with 30 c. cm. of water 
and 20 c. cm. of concentrated (sp.gr., 1.42) nitric acid until it is all dissolved (Ei). 
Test the solution for mercurous ions by diluting one drop of it in a test-tube 
with half a cubic centimeter of water and adding a drop of dilute hydrochloric 
acid (E 2 ). If mercurous ions are found to be present, add 10 c. cm. more of 
concentrated nitric acid, and boil (E 3 ) until the test with hydrochloric acid 
shows complete oxidation of the mercury. 

Dilute this solution to a volume of one liter; if a basic nitrate precipitates, 
add enough nitric acid to redissolve it. Add three drops of a molar ferric chlor- 
ide solution, and then add gradually, with constant stirring, a solution of 25 grams 
of potassium thiocyanate in 200 c. cm. of water (E 4 ), until a permanent red color 
appears. Collect the precipitate on a suction filter, dry it upon an unglazed 
plate, and place it in a dry, weighed, and labeled bottle. 

Test. Optional. After the preparation has been examined and accepted 
by the instructor, it may be made into the so-called " Pharaoh's serpent eggs" 
by kneading it to a stiff paste with a solution of 1.5 grams of dextrine in 5 c. cm. 
of water, molding this into small pellets, or cones, and drying in the air. 

Outside Questions. 

(1) What do you conclude regarding the tendency of ferric thiocyanate to 
dissociate? 

(2) How is potassium thiocyanate prepared? 

(3) From what is said in the Discussion regarding the tendency of Hg ++ to 
form complex ions, what would you expect to happen, other than simple 
dissociation, when corrosive sublimate is dissolved in water? What effect 
would this have on the apparent degree of dissociation of the corrosive 
sublimate? Is this effect actually observed? 



136 



XI. Mercuric Thiocyanate. 



LABORATORY NOTES 



DATE 







- 


Materials 
used 


Grams 


Equations for Reactions Involved in the Preparation and Tests 






- 




















Actual Yield 







Theoretical 
Yield 






Impurities 


Amount 




















t 


















Answers to Questions 



















































XII. Chloro Pentammine Cobaltic Chloride. 

j 

Suggested Reading. Roscoe and Schorlemmer, Vol. II, p. 1261. Werner, 
New Ideas on Inorganic Chemistry. 

Discussion. Both divalent and trivalent cobalt ions combine with am- 
monia to form complex ions similar to the familiar ammonio-cupric ions. A 
great variety of such complex ions of cobalt have been prepared and studied. 
The various compounds derived from them are known collectively as the cobalt- 
ammines. With the trivalent cobalt ion, when the ammonia is concentrated and 
hot, the ion [Co(NH 3 ) 6 ] +++ is the chief product; 

CQ+++ + 3Cr + 6NH 3 = [Co(NH 3 ) 6 ]+++ + 3C1~. 

At lower temperatures and at lesser concentrations of ammonia the ions [Co(NH 3 ) 5 
H 2 0]+++, [Co(NH 3 ) 4 (H 2 0) 2 ]+++, [Co(NH 3 ) 3 (H 2 0) 3 ] +++ , [Co(NH 3 ) 2 (H 2 0) 4 ]+++, 
and [Co(NH 3 ) (H 2 0) 5 ] +++ predominate; 

GO+++ + 3Cr + 5NH 3 + H 2 = [Co(NH 3 ) 5 H 2 0]+++ + 3CT, etc. 

In every case the total number of groups attached to the cobalt atom in the 
complex ion is six. This number, somewhat analogous to a valence, is charac- 
teristic of practically all cobaltammines and, indeed, of a great number of com- 
plex ions of many metals. It is called, after Werner, the coordination number 
of the cobalt atom, or of whatever other atom forms the nucleus of the complex. 

Many other neutral groups, such as hydrazine, N 2 H 4 , and hydroxylamine, 
NH 2 OH, can add on to the cobaltic ion in the same way, but the total number 
of groups is six, and since the added groups are not electrically charged, the 
charge, or valence, of the complex ion is the same as that of the original cobalt 
ion. 

The complex ions, and the salts derived from them, are named according 
to the number and nature of the groups they contain: thus, [Co(NH 3 ) 6 ] +++ 
and [Co(NH 3 ) 6 ]Cl 3 are called the hexammine cobaltic ion and hexammine 
cobaltic chloride; [Co(NH 3 ) 4 (H 2 O) 2 ]+++ and [Co(NH 3 ) 4 (H 2 O) 2 ](NO 3 ) 3 , thediaquo 
tetrammine cobaltic ion and diaquo tetrammine cobaltic nitrate, etc. 

The complex ions of trivalent cobalt are usually very stable, so that they 
give none of the reactions of the constituent groups; they do not, for instance, 
smell of ammonia, lose water vapor easily, nor precipitate cobaltic hydroxide 
with sodium hydroxide, as do ordinary cobaltic salts. 

It is possible to replace the water, the ammonia, or other neutral groups in 
these complex ions by acidic ions. Water is the most easily displaced; thus, 

137 



LABORATORY MANUAL OF GENERAL CHEMISTRY 



[Co(NH 3 ) 5 (H 2 0) ]+++ + Na++ N0 2 ~ ; [Co(NH 3 ) 5 (N0 2 ) ]++ + Na+ + H 2 

Co(NH 3 ) 4 (H 2 0) 2 ]+++ + 2Na+ + 2NOr ^ [Co(NH 3 ) 4 (NO 2 ) 2 ] + + 2Na+ + 2H 2 

Co(NH,),(H,0),]+++ + 3Na+ + 3N(V ^ [Co(NH 3 ) 3 (NO 2 ) 3 ] + 3Na+ + 3H 2 O 

Co(NH 3 ) 2 (H 2 0) 4 ]+++ + 4Na+ + 4NOr Z [Co(NH,) 2 (NO,) 4 r + 4Na+ + 4H 2 

[Co(NH 3 ) (H 2 0) 5 ]+++ + 5Na+ + 5NO 2 ~ ^ [Co(NH 3 ) (N0 2 ) 5 r ~+ 5Na+ + 5H 2 O 

[Co(H 2 0) 6 ]+++ + 6Na+ + 6NOr ^ [Co(NO 2 ) ,]- + 6Na+ + 6H 2 

The valence, or charge, of the resulting complex ions changes with each negative 
ion inserted. Each substitution decreases by one the original three positive 
charges, until a neutral, uncharged substance, Co(NH 3 ) 3 (N0 2 ) 3 , cobaltic trini- 
trito triammin^e, is formed. Further substitution adds negative charges to the 
complex ion until finally a trivalent anion is produced, [Co(N0 2 ) 6 ]~ ~, the 
hexanitrito cobaltic ion. The cobalt trinitrito triammine compound, being un- 
charged and undissociated, should not conduct electricity when in solution. 
Investigation has shown that it does not. 

There is a great difference in the ease with which different negative ions can 
be inserted in the cobaltammine complex, and in the firmness with which, once 
in, they are retained. In general, the anions of weak acids enter most readily 
and are held most tenaciously. However, the removal of an anion from a com- 
plex ion is a slow reaction, so that with the anion of even so strong an acid as 
hydrochloric contained in the complex, no precipitate is obtained for some time 
when silver nitrate is added to a freshly prepared solution. Thus chloro pen- 
tammine cobaltic nitrate, [Co(NH 3 ) 5 Cl](NO 3 ) 2 , when first dissolved in water, 
gives no precipitate with silver nitrate. Chlorine outside of the complex is, of 
course, instantly precipitated, as with any chloride. Thus with nitrito pentam- 
mine cobaltic chloride, silver chloride is at once formed; 

[Co(NH 3 ) 5 N0 2 ]Cl 2 + 2AgN0 3 = [Co(NH 3 ) 5 N0 2 ](N0 3 ) 2 + |2AgCl. 
In the preparation of chloro pentammine cobaltic chloride, hexammine co- 
baltous chloride is first formed by dissolving cobaltous chloride in a solution of 
ammonia, and then oxidized in the presence of carbonate ions; 

CoCl 2 + 6NH 3 = [Co(NH 3 ) 6 ]Cl 2 , 
[Co(NH 3 ) 6 ]Cl 2 + (NH 4 ) 2 C0 3 = [Co(NH 3 ) 6 ]C0 3 + 2NH 4 C1, 

4[Co(NH 3 ) 6 ]C0 3 + 2 + 2H 2 = 4[Co(NH 3 ) 6 ]^ 

OU 3 

Under these conditions the carbonate ion and water both replace ammonia in 
the complex, forming as the two chief products, basic carbonato tetrammine 
and basic aquo pentammine carbonate; 

OTT 

[Co(NH 3 ) 6 ] C()3 = [Co(NH 3 ) 4 C0 3 ]OH + 2NH 3 , 



and ;Co(NH 3 ) 6 ] + H 2 = [Co(NH 3 ) 5 H 2 0]^ + NH 3 . 

138 

, 



XII. CHLORO PENTAMMINE COBALTIC CHLORIDE 

Both of these compounds on being heated with dilute ammonia and ammonium 
chloride, in the absence of carbonate, are converted quantitatively into basic 
aquo pentammine chloride; 

[Co(NH 3 ) 4 C0 3 ]OH + NH 3 + 2NH 4 C1 + H 2 = [Co(NH 3 ) 5 H 2 0] H + (NH 4 ) 2 C0 3 , 






2NH 4 C1 = [Co(NH 3 ) 5 H 2 0] + (NH 4 ) 2 C0 3 . , 

If the solution is now acidified with hydrochloric acid, and heated with a large 
excess of either hydrochloric acid or some other chloride, the basic salt is con- 
verted into the neutral salt, and the water in the complex is replaced by the 
chlorine ion; 



H,0]~ + HC1 = [Co(NH 3 ) 6 H 2 0]Cl 3 + H 2 0; 
then 

[Co(NH 3 ) 5 H 2 0]Cl 3 = i [Co(NH 3 ) 5 Cl]Cl 2 + H 2 O, 
the desired product, chloro pentammine cobaltic chloride, precipitating out. 

Procedure. Dissolve 10 grams of cobalt carbonate in as little hydrochloric 
acid as possible. Filter, add 125 c. cm. of 10% aqueous ammonia, and then a solu- 
tion of 25 grams of ammonium carbonate in 125 c. cm. of water. Oxidize the 
cobaltous hexammine carbonate thus formed by sucking air rapidly through the 
solution for two or three hours, using the water suction pump, or aspirator. 
Then add 75 grams of ammonium chloride, and evaporate the solution on the 
steam bath to a slush of crystals. Add hydrochloric acid carefully until the 
solution is acid to litmus, stirring vigorously to drive off all the carbon dioxide; 
then make the solution faintly alkaline with ammonia, and add 10 c. cm, of 
concentrated aqueous ammonia in excess. Dilute the solution to 250 c. cm., 
transfer to a 500 c. cm. flask, and digest on the steam bath for one hour. Add 
150 c.cm. of concentrated hydrochloric acid, and continue the digestion for a half 
or three-quarters of an hour. Cool the solution, and decant it on to a suction 
filter, along with the precipitate of chloro pentammine chloride, separating both 
from the heavier, white crystals of ammonium chloride usually present. A little 
hydrochloric acid diluted with two volumes of water may be used to recover all 
of the pentammine chloride. Wash the precipitate on the filter with further 
small quantities of diluted hydrochloric acid until a little of the filtrate, rendered 
alkaline with sodium hydroxide and gently warmed, gives no odor of ammonia. 
Finally, wash the precipitate on the filter several times with alcohol, dry it in an 
air bath at a temperature between 100 and 125, and place in a dry, weighed, 
labeled, and stoppered bottle. 

Tests. Shake some chloro pentammine chloride with 50 c. cm. of warm 
water and divide the clear solution into five portions, using one portion for each 
of the following tests. 

139 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

(1) Boil for a few minutes. 

(2) Add a little sodium hydroxide solution and warm. 

(3) Add a little sodium sulphide solution. 

(4) Add dilute silver nitrate solution until no further precipitate occurs (Ei), 
filter, add a few drops more of the silver nitrate solution and heat (E 2 ). 

(5) Make ammoniacal, warm for a few moments (E 3 ), acidify with nitric 
acid, add an excess of silver nitrate, filter and heat. 

Add ammonium hydroxide solution gradually to a solution of a nickel salt 
(E 4 ), shake the solution with air, and add hydrogen peroxide solution. 

Outside Questions. 

(1) What conclusions do you draw from tests (l)-(3) ? From (4) ? From 
(5)? 

(2) Write the formulas of 

(A) dichloro tetrammine cobaltic nitrate; 

(B) bromo diaquo triammine cobaltic chloride; 
((7) sulphato diaquo diammine cobaltic sulphate; 
(Z>) chloro aquo tetrammine cobaltous chloride. 

(3) Write the formulas of chloro nitrato tetrammine cobaltic nitrate, and 
dinitrato tetrammine cobaltic chloride. Tell how you would distinguish 
chemically between them. 

(4) Would the sulphato tetrammine or the sulphito tetrammine ion retain its 
acid group more firmly ? 

(5) Devise and test a means for converting chloro pentammine cobalt chloride 
into chloro pentammine cobalt nitrate. 

(6) Compare cobalt and nickel as regards their tendencies to form ammines 
in the higher stage of oxidation. 



140 



XII. Chloro Pentammine Cobaltic Chloride, 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



XIII. Potassium lodate. 

Assigned Reading. Smith, pp. 277-280. 

Suggested Reading. Roscoe and Schorlemmer, Vol. II, pp. 356-364. 

Discussion. -- The stabilities of the oxygen compounds of the halogens de- 
ease m the order, iodine, bromine, chlorine, fluorine. In spite, however, of the 
relatively great stability of iodine pentoxide and of iodic acid, iodine will not 
combine directly with oxygen any more than will the other halogens. Indirect 
combination may nevertheless be brought about in many ways. Thus iodine, 
like the other halogens, reacts rapidly and reversibly with water, according to the 
equation, 

I 2 + H 2 HIO + HI. 

The balance lies far to the left in acid, and far to the right in alkaline solutions, as 
explained in Preparation No. VI, p. 1 17. The resulting hypoiodous acid oxidizes 
to the pentavalent condition even more readily than do the corresponding 
acids of chlorine and bromine, according to the equation, 

3HIO = HIO 3 + 2HI. 
Iodine may also be oxidized conveniently by concentrated nitric acid; 

5HN0 3 +I 2 + H 2 - 2HI0 3 + 5HN0 2 . 

It is simplest, unless one requires the free acid rather than its salts, to oxidize 
iodine with a chlorate; 

5KC10 3 + 31. + 3H 2 = 5KI0 3 + HI0 3 + 5HC1. 

A neutral chlorate, or in other words the chlorate ion, oxidizes iodine very 
slowly. Chloric acid acts more rapidly, and is also in all probability a somewhat 
stronger oxidizing agent than the chlorate ion, just as hypochlorous acid is a 
stronger oxidizing agent than the hypochlorite ion. Some strong acid is there- 
fore added to the chlorate solution to insure the presence of chloric acid. Since, 
however, acid is formed in the reaction itself, only a trace need be added to start 
the reaction going. 

Procedure. Dissolve 8 grams of potassium chlorate in 40 c. cm. of hot 
water in a 500 c. cm. flask. Add 10 grams of finely powdered iodine and one 
cubic centimeter of 6 molar nitric acid. Close the mouth of the flask with a 
small inverted beaker and warm the flask over one of the smaller openings on the 
steam-bath. Under these conditions of concentration and temperature the iodine 
should dissolve slowly without the evolution of any chlorine gas, no iodine vapor 
should escape from the flask, and no potassium chlorate should separate out. 
When the iodine on the bottom of the flask has completely dissolved, rinse down 
with a little water from a wash-bottle that which will have deposited in the neck 
of the flask, and warm until this too has dissolved. Transfer to a small covered 

141 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

beaker, and boil until the odor of chlorine has wholly disappeared. If, during 
this process, so much evaporation occurs that crystals of potassium iodate appear, 
add enough water to redissolve them. While still warm, neutralize carefully, 
best in two portions, with a 3 molar solution of potassium hydroxide. Cool 
thoroughly in cold water or ice, separate the crystals from the mother liquor 
by suction, and evaporate this further for a second crop. 1 Purify the entire 
product by recrystallizing from hot water, filtering the hot solution, if necessary, 
from any insoluble residue. Dry the crystals in a hot-air bath at 100-125, and 
place in a weighed and stoppered bottle. 

Tests. 

(1) Test for chloride in your preparation. This may be done by convert- 
ing the iodate into iodide with sulphur dioxide (from the cylinder) (E t ), 
boiling off the- excess of sulphur dioxide, adding silver nitrate until no 
further precipitation occurs, followed by ammonia until the liquid smells 
strongly of it (E 2 ), filtering and acidifying with nitric acid (E 3 ). 

(2) Optional. The " iodine clock." Prepare one liter of a .01 molar solution 
of potassium iodate, and 900 c. cm. of a 0,004 molar solution of sodium 
sulphite. 2 To the second solution add 5 c.cm. of molar sulphuric acid, 
and 9.5 grams of starch which has first been rubbed to a paste with a little 
cold water and then poured into 100 c. cm. of boiling water. Measure 
out 100 c. cm. of each solution into separate beakers, mix these two por- 
tions at a given instant, and count the seconds until the blue color flashes 
out. Check your results by repetition. Then dilute portions of each of 
the original solutions in various ratios and again determine the times re- 
quired for the blue to appear. Temperature differences have a marked 
effect on the rate of this reaction. 

Outside Questions. 

(1) What were the relative numbers of molecules of nitric acid, potassium 
chlorate, and iodine used in the above preparation ? 

(2) How many cubic centimeters of the 2 molar potassium hydroxide would 
you expect to use in the above neutralization ? 

(3) Explain the formation of free chlorine when the above potassium iodate 
solution is boiled (E 4 ). 

(4) From the statements in the Discussion, what would be the effect of 
adding silver acetate in excess on the equilibrium between water and 
iodine ? 

Write equilibria equations demonstrating this. 

(5) Devise a method for obtaining free iodic acid from potassium iodate, 
and write equations for the reactions you employ. 

1 The solubility of KIO, at 0, 20, 80, and 100 = 4.7, 8.1, 24.9, and 32.3 grams respectively per 
100 grams H 2 0. 

2 Sodium sulphite contains seven molecules of water of crystallization. 

142 



XIII. Potassium lodate. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



XIV. Barium Dithionate. 

Suggested Reading. Roscoe and Schorlemmer, Vol. I, pp. 447-457. 

Discussion. --The polythionic acids have the general formula H 2 Sn0 6 , 
where n is any integer from 2 to 6. As yet they have been obtained only in dilute 
aqueous solution, but many pure salts derived from them have been prepared. 
They are all strong acids; that is, their aqueous solutions react strongly acid, and 
they are excellent conductors of electricity. These solutions cannot, however, 
be concentrated beyond a certain point, as decomposition then sets in, hexa- 
thionic acid being the most sensitive, and di- and tetrathionic acids the least 
sensitive to this treatment. . The products of this decomposition are sulphuric 
acid and sulphur dioxide in the case of dithionic acid, sulphuric acid, sulphur 
dioxide, and free sulphur in the case of the higher members of the series. 

H 2 S 2 6 = H 2 S0 4 + S0 2 , 
H 2 S 3 O 6 = H 2 SO 4 + S0 2 + S, 
H 2 S 4 6 = H 2 S0 4 + SO 2 + 2S, 
etc. 

Hydrogen sulphide reduces the tetra-, penta-, and hexathionic acids to water and 
sulphur, but is without action on di- and trithionic acids. 

H 2 S 5 6 + 5H 2 S = 6H 2 + 10S, 
, etc. 

The neutral salts of all of the polythionic acids are very soluble in water. 

Dithionic acid may be prepared by the oxidation of sulphurous acid with a 
number of different mild oxidizing agents such as ferric, cobaltic, and manganic 
hydroxides, and manganese dioxide. Some sulphate is formed at the same time. 
That is, the two reactions, 

Mn0 2 + 2S0 2 = MnS 2 6 , 
and Mn0 2 + SO 2 = MnSO 4 , 

occur simultaneously. If barium hydroxide is added in excess to the resulting 
solution of manganese salts, both the sulphate and manganese ions are precipi- 
tated. Barium dithionate may be obtained by evaporation of this solution, after 
the excess of barium hydroxide has been removed; or a solution of free dithionic 
acid may be secured by adding just enough sulphuric acid to precipitate the 
barium completely as barium sulphate, and filtering. 

Procedure. Arrange a sulphur dioxide generator consisting of a dropping 
funnel and a delivery tube fitting into the neck of a 500 c. cm. flask. Place 100 

143 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

c. cm. of commercial sodium bisulphite liquor in the flask, and half fill the 
dropping funnel with concentrated sulphuric acid. 

Place 20 grams of very finely powdered manganese dioxide in a 250 c. cm. 
flask, and add 125 c. cm. of water. Surround the flask with ice water, and pass in 
a current of sulphur dioxide until nearly all of the manganese dioxide has gone 
into solution. Pour the mixture into a liter flask, add 500 c. cm. of water, warm 
to the boiling point, and add a concentrated solution containing 80 grams of 
crystalline barium hydroxide [Ba(OH) 2 2H 2 O] until no further precipitation 
occurs. Filter a small sample by suction, and test the filtrate for manganese (Ei). 
If the precipitation is incomplete, add more barium hydroxide to the hot solu- 
tion, and again test a small sample for complete precipitation. When precipita- 
tion is complete, filter, and remove the excess of barium hydroxide by passing 
carbon dioxide into the hot liquid until no further precipitation of barium car- 
bonate occurs. Filter by suction, evaporate to half the original volume, filter 
again if necessary, . and allow the solution to cool. Collect the crystals on a 
suction filter, and evaporate for another crop. Rinse the crystals with a little 
alcohol, dry, and place in a weighed and labeled bottle. 

Tests. Dissolve a gram of the barium dithionate in 30 c. cm. of cold water, 
and divide the solution into three portions. The salt should dissolve to a clear 
solution, showing the absence of barium carbonate, and the resulting solution 
should react neutral to litmus, showing absence of barium hydroxide. 

(1) Acidify one portion with hydrochloric acid and heat for a moment. 

(2) Acidify another portion with concentrated nitric acid and heat (E 2 ) . 

(3) Acidify the third portion with dilute nitric acid and add a little potassium 
permanganate (E 3 ). 

Outside Questions. 

(1) How would you identify di- and tetrathionic acids in a solution which 
might contain either or both of them ? 

(2) How many cubic centimeters of 90% sulphuric acid (sp. gr., 1.82) would 
be needed to set free all the dithionic acid from your barium dithionate ? 

(3) Optional. Write two equations for reactions which indicate that in 
tetrathionic acid the four sulphur atoms are arranged in a chain. 



144 



XIV. Barium Dithionate. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



XV. Chromous Acetate. 

Assigned Reading. - Roscoe and Schorkmmer, pp. 1017-1019. 
Suggested Reading. Roscoe and Schorlemmer, pp. 1027-1030. 

Discussion. Chromium in the divalent, chromous state has a strong ten- 
dency to go over to the trivalent, chromic state. For instance, chromous oxide, 
CrO, ignites if merely rubbed in a mortar, burning to green chromic oxide, 
Cr 2 O 3 . Dry chromous salts are slowly oxidized to chromic - salts in the air, 
while chromous salts in aqueous solutions decompose water slowly and acids 
rapidly, liberating hydrogen, and forming either basic or neutral chromic salts. 

2Cr ++ + 2H + = 2Cr +++ + H 2 . (1) 

Therefore in neutral solutions, 

2CrCl 2 + 2H 2 = 2CrCl 2 OH + H 2 ; (2) 

while in acid solutions, 

2CrCl 2 + 2HC1 = 2CrCl 3 + H 2 . (3) 

Such a behavior of a dissolved substance is almost unique, and shows that the 
chromous ion must be a very strong reducing agent. It is, indeed, the strongest 
ionic reducing agent which has so far been measured. It acts more slowly, but 
it will generate hydrogen from dilute acids with a pressure as great as metallic 
iron can produce. 

The moist acetate appears to be the most suitable form in which to preserve 
divalent chromium. In contrast with most acetates, it is sparingly soluble, and 
is therefore easy to prepare. As the salt of a weak acid, it is soluble in strong 
acids, and so is easy to use. More important still is the fact that it decomposes 
water more slowly than do most chromous salts. 

The reasons for this slow decomposition can be seen by an examination of 
equation (1). According to it the velocity of the decomposition will be propor- 
tional to the concentrations of both the chromous ion and the hydrogen ion. The 
former is small by reason of the insolubility of the acetate; the latter is small as 
a result of the hydrolysis of the acetate, due to the relative weakness of acetic acid. 

The cheapest and most convenient laboratory source of chromium is potas- 
sium dichromate. This may be reduced to chromic chloride by heating with con- 
centrated hydrochloric acid, or by the action of a great variety of reducing agents 
in the presence of dilute hydrochloric acid. The former method is unsuitable 
because of the chlorine evolved, so that the latter method, using alcohol as a re- 
ducing agent, is used here. 

145 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

The chromic chloride thus formed can exist in four modifications, the 
structural formulas and distinctive properties of which are summarized in the 
following table. 



No. 


Empirical 
Formulas 


Structural Formulas 


Color 


No. of 
Chlorine 


Fraction of 
Chlor. precip. 


No. of 
detachable 










Ions 


by AgNOj 


water mola. 


1 


CrCl,6H s O 


[Cr(H 2 0) 6 ]Cl s 


blue 


3 


3/3 





2 


CrCl 8 6H 2 O 


[Cr(H 2 0) 6 (Cl) ]C1 2 + H 2 


green 


2 


2/3 


1 


3 


CrCl s 6H 2 O 


[Cr(H 2 0) 4 (Cl) 2 ]Cl + 2H 2 


green 


1 


1/3 


2 


4 


CrCl s 


[CrCls] 


purple 












The table brings out the close similarity between these compounds and the 
cobaltammines studied in Preparation XII. Water molecules are here joined to 
the chromium atom in the same way that ammonia molecules were there joined 
to the cobalt atom. The substitution of a chlorine ion for a water molecule in the 
complex produces here the same change in valence as occurred there. 

Procedure. Pulverize 25 grams of potassium dichromate in a mortar, and 
dissolve by warming in 100 c. cm. of water, using a 500 c. cm. flask. Cool to 
room temperature under the tap, and add 75 c. cm. of concentrated hydrochloric 
acid. Place the flask under the hood, add 12 c. cm. of alcohol, and mix by shak- 
ing (Ei). Let the mixture stand until the action is over, evaporate by boiling 
to one-half the original volume, and cool under the tap. Potassium chloride 
will separate out at this point. 

Meanwhile construct an apparatus similar to a wash-bottle, using the 500 
c. cm. flask. Attach a short piece of rubber tubing to the upper end of the short 
glass tube, and attach a straight calcium chloride tube, packed loosely with a little 
asbestos or glass wool, to the top of the long glass tube. Bend the lower end of 
this long tube so that it will reach into the lowest corner of the flask. Attach 
the small end of the calcium chloride tube to a third glass tube reaching to 
the bottom of a liter flask. Be sure that your stopper and tubes are well fitted, 
so that if a gas is generated in the wash-bottle, you can force the liquid in the 
wash-bottle over into the second flask merely by pressing the rubber tip on the 
short glass tube. 

Charge the second flask with a solution of 100 grams of sodium acetate in 500 
c. cm. of water. Filter the cold solution of chromic chloride from the potassium 
chloride through a plaited filter into the first flask, or filter by suction and trans- 
fer, saving 1 c. cm. for a later test. Add 100 c. cm. of concentrated hydrochloric 
acid. Finally, add 70 grams of granulated zinc to the mixture, and replace the 
stopper with its tubes. The shorter tube will provide an outlet for the hydrogen 
gas. 

When the color of the solution in the first flask has become a clear sky-blue 
(much like that of copper sulphate solution), but while hydrogen is still being 

146 



XV. CHROMOUS ACETATE 

rapidly evolved, plug the rubber tip of the short tube, and thus force the solution 
over into the sodium acetate solution. Wash the red precipitate by decantation 
with water freed from air by bubbling a current of carbon dioxide through it, and 
preserve as a paste in a tightly stoppered bottle. 

Tests. 

(1) Dilute the 1 c. cm. of chromic chloride solution saved for this purpose 
with 30 c. cm. of water. Divide in three portions. Boil one portion for 
half a minute; then cool and compare the color with that of the other 
portion (E 2 ). Treat the second portion with a slight excess of dilute 
sodium hydroxide solution, then with a slight excess of hydrochloric acid, 
and compare this color with that of the other portions. 

(2) Dissolve a little of the acetate in dilute, air-free hydrochloric acid (E 3 ), and 
divide into several portions. 

(a) To one add a little boiled sodium hydroxide solution (E 4 ). 

(6) To another add a few drops of a solution of sodium carbonate (E 6 ). 

(c) Add another portion to a little copper sulphate solution (E e ). 

(d) Shake a fourth portion with air (E 7 ) . 

Outside Questions. 

(1) If the alcohol were oxidized by the dichromic acid to acetic acid, how many 
cubic centimeters of 95% alcohol (sp.gr., 0.81) would be required to 
reduce to chromic chloride, in dilute hydrochloric acid, the potassium 
dichromate which you took ? 

(2) What is the coordination number of chromium in CrCl 3 6H 2 O ? 

(3) What would be the name of the third variety of chromic chloride given in 
the table ? 

(4) How do you explain what happened when alkali and then acid were added 
to the solution of green chromic chloride ? 



147 



XV. Chromous Acetate. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



XVI. Potassium Thiocarbonate. 

Suggested Reading. Roscoe and Schorlemmer, Vol. I, p. 813. 

Discussion. The marked similarity of the elements oxygen and sulphur 
is strikingly illustrated by the action of alkali oxides on carbon dioxide on the one 
hand, and of alkali sulphides on carbon disulphide on the other. 

Na 2 O + C0 2 = Na 2 CO 3 , 
Na 2 S + CS 2 = Na 2 CS 3 . 

The two reactions are precisely analogous; the one gives a carbonate, the other a 
thiocarbonate. 

Thiocarbonic acid is formed by the action of any strong acid on a thiocar- 
bonate. It is a yellow, evil-smelling oil, slightly soluble in water, and when 
warmed decomposes into carbon disulphide and hydrogen sulphide. It is evi- 
dently a little more stable than carbonic acid, for that substance decomposes so 
readily into carbon dioxide and water that it has never been obtained in the free 
state. 

H 2 CS 3 = Co 2 -f- H 2 S, 
H 2 CO 3 = CO 2 + H 2 O. 

Procedure. Saturate 25 c. cm. of molar potassium hydroxide solution with 
hydrogen sulphide (Ei), and mix this with an equal volume of the untreated 
potassium hydroxide solution (E 2 ) (Q t ). Shake this solution for five minutes 
with 2 c. cm. of carbon disulphide, and then pour it through a filter previously 
moistened with water (Q 2 ). 

Prepare a small quantity of a 10~ 4 molar solution of any nickel salt. Mix 
1 c. cm. of the thiocarbonate solution with 2 c. cm. of this solution. Then, by 
successive dilution, prepare small quantities of 10~ 5 , 10~ 6 , etc. molar solutions of 
the nickel salt until 2 c. cm. of the solution gives no color change when mixed 
with 1 c. cm. of the thiocarbonate solution. Determine the extent to which 
cobalt salts interfere with this test by making comparative tests with solutions 
containing 1, 2, 10, 50, etc. parts of cobalt to one of nickel. 

Tests. Add a few drops of the thiocarbonate solution to a few cubic 
centimeters of a solution of (1) silver nitrate (E 3 ), (2) barium chloride (E 4 ), 
(3) lead nitrate (E 5 ), and allow the resulting solutions to stand for a few minutes. 

Outside Questions. 

(1) What advantages has this method of preparing the sulphide ? 

(2) Why was the filter paper previously wetted ? 

(3) Explain the color changes which occur in Tests (1) and (2). 

149 



XVI. Potassium Thiocarbonate. 



LABORATORY NOTES 



DATE. 



Materials 
used 



Grams 



Equations for Reactions Involved in the Preparation and Tests 



Actual Yield 



Theoretical 
Yield 



Impurities 



Amount 



Answers to Questions 



XVII. Selenium and Tellurium from Tellurium Slag. 

Assigned Reading. Smith, pp. 401-404. 

Suggested Reading. Roscoe and Schorlemmer, Vol. I, pp. 458-484. 

Discussion. In the electrolytic refining of copper the metallic impurities 
baser than copper in the anode plates dissolve along with the copper. The 
metallic impurities nobler than copper do not dissolve, and as the copper anodes 
are eaten away, fall to the bottom of the cell as a fine powder. Mixed with 
this powder are other insoluble materials either originally contained in the 
anodes themselves or formed from them, such as selenium and tellurium, lead 
sulphate, silica, basic sulphates of arsenic, antimony, and bismuth, and particles 
of copper which have become detached from the anodes in the process of solution. 
This mixture, accumulating under the anodes, is called the anode slime, or 
anode mud. 

The refiner as a rule is chiefly interested in the gold and silver in this anode 
slime. To recover these metals, the slime is first digested with hot, aerated 
sulphuric acid, which eliminates most of the copper, arsenic, antimony, and 
bismuth as sulphates. The residue is then fused in a furnace with lead and a 
flux of sodium nitrate and sodium carbonate. The selenium and tellurium are 
oxidized by the sodium nitrate, and with the sodium carbonate form a fluid slag 
of sodium selenate and sodium tellurate. The slag thus produced is drawn off 
from the molten lead and allowed to solidify. The gold and silver dissolve in 
the molten lead. This alloy is drawn off, and the lead eliminated as lead oxide 
by heating in the air. 

This so-called tellurium slag, consisting chiefly of sodium selenate and 
tellurate with small amounts of lead, copper, silver, and gold as impurities, 
is a convenient source of selenium and tellurium. It dissolves readily in 
concentrated hydrochloric acid, giving a solution from which free selenium 
and tellurium can be precipitated by a variety of reducing agents. 

Procedure. -- To 25 grams of finely powdered slag in a 500 c. cm. flask add, 
under the hood, 120 c. cm. of concentrated hydrochloric acid (Ei). Let stand 
with frequent shaking until no more chlorine is evolved and all the slag has dis- 
appeared. Dilute with 100 c. cm. of water, and filter from silver chloride and 
any black residue, collecting the filtrate in a 500 c. cm. Erlenmeyer flask. Cool 
this strongly acid solution of selenium and tellurium chlorides to about zero 
degrees in a salt-and-ice bath, and add, under the hood, 60 grams of sodium 
bisulphite, a few grams at a time, shaking vigorously after each addition and not 

151 



LABORATORY MANUAL OF GENERAL CHEMISTRY 

allowing the temperature to rise above 10 (E,). 1 When all of the bisulphite has 
been added, and the evolution of sulphur dioxide has ceased, stopper the flask 
and let it stand in the ice bath for half an hour with occasional shaking. Then 
filter by suction, keeping the mixture cold until it is filtered. To make sure 
that all the tellurium has been precipitated, dilute with 20 c. cm. of water and 
add a few grams more of the bisulphite. Collect any further precipitate and 
add it to that already obtained. Wash the combined precipitate a few times 
with somewhat diluted hydrochloric acid, and then a few times with cold water. 
Transfer the mixture of selenium and tellurium to a casserole, and dissolve 
it in as little concentrated nitric acid as possible, warming to hasten the action 
(E,). Evaporate cautiously to 5 c. cm. on a wire gauze over a flame and then 
to dryness on the steam bath. To complete the removal of the nitric acid, add 
5 c. cm. of concentrated hydrochloric acid and evaporate to dryness. Repeat 
this treatment. 

Dissolve the residue in 23 c. cm. of water and 145 c. cm. of concentrated 
hydrochloric acid in a 500 c. cm. flask, and treat with 25 grams of sodium bi- 
sulphite just as before, taking especial precautions to keep the temperature low. 
Allow the solution to stand for a half hour with frequent shaking, and then filter 
by suction. The filtrate should be yellow or green and should smell strongly of 
sulphur dioxide. If it is red, it contains colloidal selenium, and this should be 
precipitated by the addition of more bisulphite with frequent shaking at a low 
temperature. Wash the combined precipitates of selenium, first with dilute 
hydrochloric acid, and then with water, until no test for chloride is obtained. 
Finally wash several times with alcohol, and dry on a porous plate away from 
any source of heat. 

Dilute the filtrate containing the tellurium chloride with an equal volume 
of water and, without cooling, add 40 grams of sodium bisulphite, a little at a 
time, shaking after each addition. Stopper and let stand for an hour. Filter, 
test the filtrate for complete precipitation of the tellurium, and wash and dry 
it as in the case of selenium. 

Heat some crude concentrated sulphuric acid to 150 in a small beaker, and 
adjust a flame under it so that this temperature will be approximately maintained. 
Place the dry, red selenium in a small test-tube, insert a thermometer so that the 
bulb is surrounded by the powder, lower the test-tube into the acid, and observe 
the temperature indicated on the thermometer. When no further change occurs 
remove the test-tube, cool, rinse off the sulphuric acid, and heat over a flame 
until the selenium melts, noting the temperature. Transfer the solidified selenium 
to a weighed and labeled bottle. 



WU1 P reci P itate both selenium and tellurium from their chlorides dissolved in 
hydrochloric acid. A sufficient excess of concentrated acid was used at the outset to give about 
that concentration at this point. In 10 M hydrochloric acid selenium only is precipitated by sulphur 
:ide. Ihe later separation of selenium and tellurium depends on this fact. 

The precipitation is carried on at a low temperature to keep the selenium in the red form, for in 
this condition it is more readily filtered off and washed. 



152 



XVII. SELENIUM AND TELLURIUM FROM TELLURIUM SLAG 
Tests. 

(1) Apply a colorless flame to small amounts of selenium and tellurium 
separately on the cover of a porcelain crucible (E 4 ). 

(2) Shake small amounts of selenium and tellurium separately with a little 
colorless concentrated sulphuric acid. 

Outside Questions. 

(1) Cite evidence from the above preparation indicating whether selenium 
or tellurium resembles the element sulphur the more closely. 

(2) Explain the temperature changes indicated by the thermometer when 
selenium was heated. 



153 



XVII. Selenium and Tellurium from Tellurium Slag. 



LABORATORY NOTES 



DATE 









Materials 
used 


Grams 


Equations for Reactions Involved in the Preparation and Tests 
























1 ' 


Actual Yield 






Theoretical 
Yield 






Impurities 


Amount 






































Answers to Questions 



















































APPENDIX 



APPARATUS REQUIRED FOR EXPERIMENTS AND EXERCISES IN 

THIS MANUAL 

1 Asbestos paper, 15 cm. square. 

1 Beaker, Griffin, 75 c. cm. 

1 Beaker, Griffin, 150 c. cm. 

1 Beaker, Griffir>, 250 c. cm. 

1 Beaker, Griffin, 350 c. cm. 

1 Beaker, Griffin, 500 c. cm. 

1 Beaker, tall, w. lip, 550 c. cm. 

1 Beaker, resistance glass, Griffin, 150 c. cm. 

1 Beaker, resistance glass, Griffin, 250 c. cm. 

1 Beaker, resistance glass, Griffin, 400 c. cm. 

1 Beaker, resistance glass, Griffin, 600 c. cm. 

1 Beaker, resistance glass, Griffin, 800 c. cm. 

1 Beaker, resistance glass, Griffin, 1000 c. cm. 

1 Boat, porcelain, 80 mm. 

1 Bottle, acid, green, 2000 c. cm. 

4 Bottles, salt mouth, g. s., 60 c. cm. 

1 Bottle, wide mouth, 250 c. cm. 

1 Box, pasteboard, for filter paper. 

1 Brush, camel's hair. 

1 Brush, test-tube. 

2 Burners, Bunsen. 

1 Casserole, porcelain, 210 c. cm. 

1 Casserole, porcelain, 750 c. cm. 

2 Clamps, iron, small. 

1 Crucible, porcelain, 25 c. cm. 

1 Crucible cover, porcelain, for 25 c. cm. crucible. 

1 Crucible, iron, 50 c. cm. 

1 Cylinder, graduated, 100 c. cm. 

1 Dish, agateware, 15 cm. 

1 Dish, evaporating, porcelain, 100 mm. 

1 File, round. 

1 File, triangular. 

1 Filter arm, 2-hole. 

1 Package filter paper, 7 cm. 

1 Package filter paper, 12 cm. 

1 Package filter paper, 15 cm. 

6 Sheets filter paper, 50 x 50 cm. 

6 Sheets, filter paper, hardened, 5.5. cm. 

1 Flask, Erlenmeyer, 250 c. cm. 

1 Flask, Erlenmeyer, 500 c. cm. 

Flask, filtering, Erlenmeyer, w. side tube, 250 c. cm. 

Flask, filtering, Erlenmeyer, w. side tube, 1000 c. cm. 

1 Flask, flat bottom, 100 c. cm. 

2 Flasks, flat bottom, 250 c. cm. 
2 Flasks, flat bottom, 500 c. cm. 
1 Flask, flat bottom, 1000 c. cm. 
1 Funnel, 65 mm. 

156 



1 Funnel, 105 mm. 

1 Funnel, porcelain, Hirsch, 100 mm. 

1 Funnel, separatory, cylindrical, 30 cc. 

2 Funnel tubes. 

2 Gauzes, iron, 15 cm. square. 

1 Jar, screw cap, for litmus paper, etc. 

2 Packages litmus paper. 

1 Mortar and pestle, porcelain, 70 mm. 

1 Pincers, iron. 

1 Pinchcock, screw. 

1 Pinchcock, spring. 

2 Plates, glass. 
1 Plate, iron. 

3 Plates, porous. 

1 Retort stand, medium, 2 rings. 

2 20 cm. Lengths rod, glass, 4 mm. 
1 Spatula, horn, small. 

1 Sponge. 

2 Stoppers, rubber, 12 mm., 1-hole, No. 0. 
2 Stoppers, rubber, 17 mm., 1-hole, No. 2. 
1 Stopper, rubber, 17 mm., 2-hole, No. 2. 
1 Stopper, rubber, 20 mm., 2-hole, No. 4. 
1 Stopper, rubber, 23 mm., 2-hole, No. 5. 

1 Stopper, rubber, 26 mm., solid, No. 6. 

2 Stoppers, rubber, 26 mm., 2-hole, No. 6. 
1 Stopper, rubber, 30 mm., 1-hole, No. 7. 

12 Test-tubes, 160 x 12 mm. 

3 Test-tubes, 160 x 25 mm. 

1 Test-tube, resistance glass, 120 x 16 mm. 

1 Test-tube rack, 1 2-hole. 

1 Thermometer, grad. on stem, 360 in 1. 

1 Tongs, iron. 

1 Towel, cotton. 

1 Triangle, pipestem, 50 mm. 

1 Tripod, iron. 

1 Tube, CaCl 2 , straight, 100 mm. 

1 Tube, U, 105 mm., 2 side tubes. 

2 Tubes, U, special. 

1 80 cm. Length tubing, glass, 4.5 mm. 

1 35 cm. Length tubing, hard glass, 18 mm. 

1 60 cm. Length tubing, rubber, 5 mm. 

3 60 cm. Lengths tubing, rubber, 6.5 mm. 

1 60 cm. Length tubing, rubber, suction, 5 mm. 

2 Watch glasses, 50 mm. 
1 Watch glass, 80 mm. 

1 Watch glass, 130 mm. 

1 Wing top, for Bunsen flame. 

3 Meters wire, copper No. 18. 

1 6 cm. Length wire, platinum, No. 27. 

2 Wooden blocks, 5 x 5 x 10 cm. 



ATOMIC WEIGHTS AND LOGARITHMS OF ATOMIC WEIGHTS 



Based on the International Atomic Weights for 1916 



ELEMENT 


SYMBOL 


ATOMIC 
WEIGHT 


LOGARITHM 
OP ATOMIC 
WEIGHT 


ELEMENT 


SYMBOL 


ATOMIC 
WEIGHT 


LOGARITHM 
OF ATOMIC 
WEIGHT 


Aluminum 


Al 


27.1 


I.432Q7 


Mercury 


Hg 


2OO.6 


2.30233 


Antimony 


Sb 


1 20. 2 


2.O7QQO 


Molybdenum 


Mo 


96.0 


1.98227 


Argon 


A 


30.88 


1.6007=; 


Neodymium 


Nd 


144.3 


2.15927 


Arsenic 


As 


74 06 


1.87483 


Neon 


Ne 


2O.2 


1.30535 


Barium 


Ba 


127.77 


2.I378Q 


Nickel 


Ni 


58.68 


1.76849 


Bismuth 


Bi 


2O8.O 


2.31806 


Nitrogen 


N 


I4.OI 


1.14644 


Boron 


B 


II.O 


I.O4I3Q 


Osmium 


Os 


190.9 


2.28081 


Bromine 


Br 


70.02 


1.90266 


Oxygen 


O 


I6.OOO 


1.20412 


Cadmium 


Cd 


II2.4O 


2.0^077 


Palladium 


Pd 


IO6.7 


2.02816 


Caesium 


Cs 


132.81 


2. 1232? 


Phosphorus 


P 


31.04 


1.49192 


Calcium 


Ca 


4O.O7 


I.6O282 


Platinum 


Pt 


195.2 


2.29048 


Carbon 


c 


I2.OO5 


I.O7Q36 


Potassium 


K 


39-IO 


1.59218 


Cerium 


Ce 


I4O.2=; 


2.I460O 


Praseodymium 


Pr 


140.9 


2.14891 


Chlorine 


Cl 


3C.46 


I. =14074 


Radium \ . . 


Ra 


226.O 


2-354 11 


Chromium 


Cr 


=12.0 


1.71600 


Rhodium . . .* 


Rh 


IO2-9 


2.01241 


Cobalt 


Co 


=;8.Q7 


I.77O63 


Rubidium 


Rb 


8545 


1.93171 


Columbium 


Cb 


Q2. cr 


I. Q7O8l 


Ruthenium 


Ru 


IOI-7 


2.00732 


Copper 


Cu 


63. =;7 


1.8032* 


Samarium 


Sa 


150.4 


2.17725 


Dysprosium 


Dy 


162 =; 


2. 2IO8* 


Scandium 


Sc 


44-1 


1.64444 


Erbium 


Er 


167.7 


2.2241:3 


Selenium 


Se 


79-2 


1.89873 


Europium 


Eu 


I C2 O 


2 18184 


Silicon 


Si 


28.3 


1.45179 


Fluorine 


F 


10. 


1.2787"; 


Silver 


Ag 


107.88 


2.03294 


Gadolinium 


Gd 


T C7 2. 


2 IQ673 


Sodium 


Na 


23.OO 


1.36173 


Gallium 


Ga 


60.0 


1.84448 


Strontium 


Sr 


87.63 


1.94265 


Germanium 


Ge 


72 =; 


1.86034 


Sulphur 


S 


32.06 


1.50596 


Glucinum 


Gl 


91 


O.Q=;QO4 


Tantalum 


Ta 


I8I.5 


2.25888 


Gold 


Au 


107.2 


2.20401 


Tellurium 


Te 


I2 7-5 


2.10551 


Helium 


He 


4.OO 


O.6O2O6 


Terbium 


Tb 


159.2 


2.20194 


Holmium 


Ho 


163 ? 


2.213 *2 


Thallium 


Tl 


204.0 


2.30963 


Hydrogen 


H 


I 008 


O OO34.6 


Thorium 


Th 


232.4 


2.36624 


Indium 


In 


II4.8 


2.O*QQ4 


Thulium 


Tm 


168.5 


2.22660 


Iodine 


I 


126.92 


2.IO3*3 


Tin 


Sn 


118.7 


2.07445 


Iridium 


Ir 


IQ3. 1 


2. 2808 


Titanium 


Ti 


48.1 


1.68215 


Iron 


Fe 


cc.84 


1.74604. 


Tungsten 


W 


184.0 


2.26482 


Krypton 


Kr 


82.92 


1.91866 


Uranium 


U 


238.2 


2.37694 


Lanthanum 


La 


I3Q.O 


2.I43OI 


Vanadium 


V 


51.0 


1.70757 


Lead 


Pb 


2O7.2O 


2.31618 


Xenon 


Xe 


130.2 


2.11461 


Lithium 


Li 


6.Q4 


0.84136 


Ytterbium 


Yb 


173.5 


2.23930 


Lutecium 


Lu 


17=;. O 


2.243O4 


Yttrium 


Yt 


88.7 


1.94792 


Magnesium 


Me 


24. 32 


1. 38=106 


Zinc 


Zn 


65.37 


1.81538 


Manganese 


Mn 


C4.Q3 


1. 73Q8l 


Zirconium 


Zr 


90.6 


^957 I 3 



















158 



PQ 



OS 



PQ 



CO 

^H rf 
ui 

CO 



CD 

co 



00 



00 

E5 



o os 
06 



Tt< 

0) 00 
fe 10 

10 





OS 



co 

a < 



PQ 



o 

a 



CO 



CD 
) OS 



o 

> ^ 

to 



? CD 

ft-O 



as 



3 ,4 
p^ Q 



as 

CD 
(M 







(M 



CO 



OS 



00 
CO 



o 

(M 



^ 

H oo 



a 
3 
o 







PQ 



o 



00 
(M 






PQ 



as 

CD 



O 

PQ ^ 





^ 



PQ 



W) co 



o 
S 



PQ 



PQ 



a 
o 
O 



c 

PL, 






o 

aS O 




O 
O 

CO 



co 
o 



cd O 
d 



SJ 



00 



CO 

CD 



00 



H co 



(N 



cj as 
i-l co 



CO 



w 



p 



Q 

CO 



00 

co 



n 
PH 



00 



CM 

00 



^ ' 

O Pi 

o 

CJ 

CO 



(N 

os 



. 
o <M 

w CO 



(M 

O 



W) 
C 

o 



LOGAEITHMS 



Nat. 





1 


2 


3 


4 


5 


6 


7 


8 


Q 


PROPORTIONAL PARTS. 


Nos. 


















\j 


\J 


123 


456 


789 


10 


oooo 


0043 


0086 


0128 


0170 


O2I2 


0253 


0294 


0334 


0374 


4 8 12 


17 21 25 


2 9 33 37 


11 


0414 


453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


4811 


1 S 19 23 


26 30 34 


12 


0792 


0828 


0864 


0899 


934 


0969 


1004 


1038 


1072 


1106 


3 7 10 


14 17 21 


24 28 31 


13 


"39 


"73 


1206 


1239 


1271 


1303 


1335 


1367 


'399 


143 


3 6 10 


13 16 19 


23 26 29 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


369 


12 15 IS 


21 24 27 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


'959 


1987 


2014 


3 6 8 


II 14 17 


20 22 25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


22OI 


2227 


2253 


2279 


3 5 8 


ii 13 16 


18 21 24 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2 455 


2480 


2504 


2529 


2 5 7 


IO 12 15 


17 2O 22 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


2 5 7 


9 12 14 


16 19 21 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


247 


9 " 13 


16 18 20 


20 


3010 


3032 


3054 


3075 


3096 


3II8 


3139 


3160 


3181 


320? 


246 


8 ii 13 


15 17 19 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


246 


8 10 12 


14 16 18 


22 


3424 


3444 


3464 


3483 


3502 


3522 


354i 


356o 


3579 


3598 


246 


8 IO 12 


14 15 17 


23 


361? 


3636 


3655 


3674 


3692 


37" 


3729 


3747 


3766 


3784 


246 


79" 


13 15 17 


24 


3802 


3820 


3838 


3856 


3874 


3892 


399 


3927 


3945 


3962 


245 


79" 


12 14 16 


25 


3979 


3997 


4014 


403 1 


4048 


4065 


4082 


4099 


4116 


4133 


235 


7 9 10 


12 14 IS 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


2 3 5 


7 8 10 


II 13 15 


27 


43M 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


235 


689 


II 13 14 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


2 3 5 


689 


II 12 14 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


i 3 4 


679 


IO 12 13 


30 


477i 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


i 3 4 


679 


10 ii 13 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


i 3 4 


678 


IO II 12 


32 


5051 


5065 


5079 


5092 


5io5 


5"9 


5132 


5145 


5159 


5172 


i 3 4 


5 7 8 


9III2 


33 


5185 


5198 


52" 


5 22 4 


5237 


5250 


5263 


5276 


5289 


5302 


i 3 4 


568 


9 10 12 


34 


5315 


5328 


534 


5353 


5366 


5378 


5391 


5403 


54i6 


5428 


i 3 4 


568 


9 10 ii 


35 


5441 


5453 


5465 


5478 


549 


552 


55H 


5527 


5539 


5551 


I 2 4 


5 6 7 


9 10 ii 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


I 2 4 


5 6 7 


8 10 ii 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


I 2 3 


5 6 7 


8 9 10 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 5888 


5899 


I 2 3 


5 6 7. 


8 9 10 


39 


59" 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


I 2 3 


457 


8 9 10 


MO 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


I 2 3 


456 


8 9 10 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


I 2 3 


456 


7 8 9 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


63H 


6325 


I 2 3 


456 


7 8 9 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


I 2 3 


456 


7 8 9 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


I 2 3 


456 


7 8 9 


45 


6532 


6542 


655i 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


I 2 3 


456 


7 8 9 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


i 2 3 


456 


7 7 8 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


I 2 3 


455 


678 


48 


6812 


68.21 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


i 2 3 


445 


678 


49 


6902 


691 1 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


I 2 3 


445 


678 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


I 2 3 


345 


678 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


I 2 3 


345 


678 


52 


7160 


7168 


7177 


7185 


7i93 


7202 


7210 


7218 ! 7226 


7235 


122 


345 


677 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


73i6 


122 


3 4 5 


667 


54 


7324 


7332 


7340 


7348 


735 6 


7364 


7372 


7380 


7388 


739 6 


I 2 2 


345 


667 



160 



Nat. 
Nos. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


PROPORTIONAL PARTS. 
























123 


456 


789 


55 


7404 


7412 


7419 


7427 


7435 


7443 


745 


7459 


7466 


7474 


I 2 


345 


567 


56 


7482 


749 


7497 


7505 


75i3 


7520 


752 


7536 


7543 


755 


122 


345 


567 


57 


7559 


7566 


7574 


7582 


7589 


7597 


760^ 


7612 


7619 


762 


122 


345 


567 


58 


7634 


7642 


7649 


7657 


7664 


7672 


767 


7686 


7694 


770 


112 


344 


J 1 

567 


59 


7709 


7716 


7723 


7731 


7738 


7745 


775 


7760 


7767 


7774 


112 


344 


5 6 7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


782 


7832 


7839 


7846 


I I 2 


344 


566 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


112 


344 


566 


62 


7924 


793' 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


I I 2 


334 


566 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


I I 2 


334 


556 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


I I 2 


334 


5 5 6 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


II 2 


334 


5 5 6 


66 


8i95 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


112 


334 


5 5 6 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


I I 2 


334 


5 5 6 


68 


8325 


833 1 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 


112 


334 


456 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


I I 2 


2 3 4 


456 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


112 


234 


456 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


112 


234 


455 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


112 


234 


455 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


112 


234 


455 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


112 


234 


4 5 5 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


112 


233 


455 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


112 


233 


455 


"7 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


112 


233 


445 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


I I 2 


233 


445 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9 OI 5 


9020 


9025 


112 


233 


445 


80 


9031 


9036 


9042 


9047 


953 


9058 


9063 


9069 


9074 


9079 


112 


233 


445 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 




I I 2 


233 


445 


82 


9 J o8 


9 J 43 


9149 


9 J 54 


9 1 59 


9 l6 5 


9170 


9175 


9180 


9186 


112 


233 


445 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


I 2 


233 


445 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


I 2 


233 


445 


85 


9294 


9299 


934 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


I 2 


233 


445 


86 


9345 


935 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


939 


I 2 


233 


445 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


943 


9435 


9440 


I I 


223 


344 


88 


9445 


945 


9455 


9460 


465 


9469 


9474 


9479 


9484 


9489 


I I 


223 


344 


89 


9494 


9499 


9504 


9509 


5i3 


95i8 


9523 


9528 


9533 


9538 


I I 


223 


344 


90 


9542 


9547 


9552 


9557 


562 


9566 


957i 


576 


958i 


9586 


I I 


223 


344 


91 


959 


9595 


9600 


9605 


609 


9614 


9619 


624 


9628 


9633 


I I 


223 


344 


92 


9638 


9 6 43 


9647 


9652 


657 


9661 


9666 


9671 


9675 


9680 


I I 


223 


344 


93 


9685 


689 


9694 


9699 


703 


9708 


9713 


717 


9722 


9727 


I I 


223 


344 


94 


9731 


736 


974i 


9745 


750 


9754 


759 


763 


9768 


9773 


I I 


2 2 3 


344 


95 


9777 


9782 


9786 


9791 


795 


9800 


805 


809 


981.4 


SiS 


I I 


223 


344 


96 


9823 


827 


9832 


9836 


841 


845 


9850 


854 


9859 


863 


Cf I I 


223 344 


97 


9868 


872 


9877 


9881 


886 


890 


9894 


899 


993 


908 


I 1 


223 344 


98 


9912 


917 


9921 


9926 


93 


934 


9939 


943 


9948 


952 


I I 


223 


344 


99 


9956 


961 


9965 


969 


974 


978 


9983 


987 


999 


996 


I I 


223 


334 



161 



LOGARITHMS OF CHEMICAL FACTORS 

Calculated from the International Atomic Weights for 1916 





WEIGHED 


RE- 


LOG. OF 




W EIGHED 


RE- 


LOG. OP 






QUIRED 


FACTOR 






QUDJED 


FACTOR 


Aluminum (27.1) 
Antimony (120.2) 


A1 2 3 
Sb 2 4 


Al 

Sb 


9-72455 
9.89749 


Magnesium (24.32) 


Mg 2 P 2 O 7 

a 


Mg 
MgO 


9-33924 
9.55879 


Arsenic (74.96) 


Sb 2 S 
As 2 S 3 


Sb 
As 


9.85384 
9-78475 


Manganese (54.93) 


Mn 3 4 
Mn 2 P 2 7 


Mn 
Mn 


9-85749 
9.58761 




a 


As 2 3 


9-90538 


Mercury (200.6) 


H g2 Cl 2 


Hg 


9.92931 




Mg2As 2 7 


As 


9.68371 




HgS 


Hg 


9.93561 


i 


U 


As 2 3 


9.80434 


Molybdenum (96.0) 


Mo0 3 


Mo 


9.82391 


Barium (137.37) 


(( 

BaSO 4 


As 2 5 
Ba 


9.86943 
9.76974 


Nickel (58.68) 


NiO 

NiS0 4 


Ni 
Ni 


9.89529 
9.57889 




a 


BaO 


9.81758 


Nitrogen (14.01) 


Pt(i95-2) 


N 


9.15699 




BaSiFe 


Ba 


9.69125 







NH 3 


9.24187 


Bismuth (208.0) 


Bi 2 3 


BaO 
Bi 


9.73909 
9-95258 


Phosphorus (31.04) 



Mg 2 P 2 7 


NH 4 
P 


9.26684 
9-445 J 9 


Bromine (79.92) 


BiAs0 4 
AgBr 


Bi 
Br 


9.77778 
9.62896 




i 

a 


P0 4 
P 2 5 


9.93118 
9.80477 


Cadmium (112.4) 


Ag 
CdO 


Br 
Cd 


9.86971 
9.94220 


Potassium (39.10) 


K 2 PtCl 6 




K 
K 2 O 


9.20643 

0.28727 


Calcium (40.07) 


CaO 


Ca 


9.85409 




KC1 


K 


;/ / ~ / 

9.71967 




CaSO 4 


Ca 


9.46886 




a 


K 2 


9.80051 


Carbon (12.005) 
Chlorine (35.46) 


it 

C0 2 

AgCl 


CaO 
C 
Cl 


9.61478 
9-43586 

9-39337 


Silicon (28.3) 


K 2 SO 4 

u 

Si0 2 


K 

K 2 O 

Si 


9.65201 
9-73285 
9.67147 







HC1 


9-40555 


Silver (107.88) 


AgCl 


Ag 


9.87657 


Chromium (52.0) 


Ag 
Cr 2 3 


Cl 
Cr 


9.51680 
9-835I9 




AgBr 
Agl 


Ag 

Ag 


9-75924 
9.66224 







CrO 3 


0.11919 


Sodium (23.00) 


NaCl 


Na 


9-59487 




fl 


CrO 4 


0.18364 




u 


Na 2 O 


9.72450 




BaCr0 4 


Cr 


9.31225 




Na 2 S0 4 


Na 


9.51029 




(I 


CrO 3 


9.59624 




u 


Na 2 O 


9.63992 




U 


Cr0 4 


9.66070 


Strontium (87.63) 


SrS0 4 


Sr 


9.67857 


Cobalt (58.97) 


It 

CoS0 4 


Cr 2 O 7 
Co 


9.62967 
9.58022 




u 
SrC0 3 


SrO 
Sr 


9.75140 
0.77346 


Copper (63.57) 


Cu 


CuO 


0.09750 




a 


SrO 


y / / \_/T^ V 

9.84630 




CuO 


Cu 


9.90250 


Sulphur (32.06) 


BaSO 4 


S 


9.13781 




Cu 2 S 


Cu 


9.90234 




a 


S0 2 


0.4^843 


Fluorine (19.0) 


CaF 2 


F 


9.68730 




u 


S0 3 


;* ~_/ T^} 
9.53526 


Hydrogen (1.008) 
Iodine (126.92) 


H 2 
Agl 


H 
I 


9.04884 
9-73283 




u 
u 


SO 4 
H 2 SO 4 


9.61439 
9.62341 


Iron (55.84) 


Ag 
Fe 2 3 


I 

Fe 


0.07059 
9-84473 


Tin (118.7) 


PbSO 4 

SnO 2 


S 
Sn 


9.02415 
9.89634 


Lead (207.2) 


U 

PbO 


FeO 
Pb 


9-954I5 
9.96770 


Titanium (48.1) 
Tungsten (184) 


TiO 2 
WO 3 


Ti 
W 


9.77852 
9.89933 




PbS 


Pb 


9-93752 


Uranium (238.2) 


U0 2 


U 


9.94526 


Lithium (6.94) 


PbSO 4 
Li 3 PO 4 


Pb 
Li 


9.83458 
9-25455 


Zinc (65.37) 


ZnO 

ZnS 


Zn 
Zn 


9.90491 
9.82669 


Magnesium (24.32) 


t( 

MgO 


Li 2 
Mg 


9-58754 
9.78044 





ZnS0 4 


ZnO 
Zn 


9.92177 
9.60740 












u 


ZnO 


9.7O248 



162 



INDEX 



ABSORPTION OF IODIDES BY THE STOMACH, 55. 

Accidents, 5, 6. 

Acetic acid, test, 72. 

Acetylene, 68. 

AFFINITIES OF THE HALOGENS FOR NEGATIV 

ELECTRICITY, 53. 
ALCOHOLS, 69. 

ALKALINE EARTH ELEMENTS, 75. 
Alums, 111. 
ALUMINUM, 83. 

AMMONIA AND AMMONIUM HYDROXIDE, 59. 
AMMONIO-CUPRIC SULPHATE, 115. 
AMMONIUM BROMIDE, 117. 
Ammonium hexachlor plumbate, 88. 
AMMONIUM IRON ALUM FROM SIDERITE, 111. 
Amphoteric substance, 121. 
ANTIMONY, 85. 
Apparatus, list of, 155. 
Aqueous tension, table, 20. 
ARSENIC ACID, 121. 

ARSENIC, ANTIMONY, AND BISMUTH, 85. 
Arsine, 85. 
Aspirator, 12, 101. 
Atomic weights, table, 158. 

AUTOXIDATION, 17. 

Balances, use of, 4. 

BARIUM, 75. 

BARIUM DITHIONATE, 143. 

Barometer corrections, table, 19. 

Baryta water, 67. 

Birkland and Eyde, 99. 

BISMUTH, 85. 

BOILING POINT, EFFECT OF DISSOLVED SOLIDS 

ON, 27. 
Borates, 73. 
Borax, 73. 
Boric acid, 73. 
Boric acid, test for, 73. 
BORON AND SILICON, 73. 
Bromic acid, 56. 

BROMINE AND IODINE, PROPERTIES OF, 49. 
Bromine burns, 118. 
Burns, treatment, 6. 

Cadmium, 79. 

Calcium, 75. 

Carbonates, alkaline earth, 75. 



CARBON MONOXIDE, 67. 
Carron oil, 6. 
CATALYSIS, 47. 
Catalysts, 47. 
Catalytic agents, 47. 
Celestite, 107. 

CHEMICAL EQUILIBRIUM, 31. 
CHEMICAL REACTION, VELOCITY OF, 29. 
Chemical supplies, 4, 97. 
Chemist's Garden, 74. 
Chloric acid, 55. 
Chloride of lime, 55. 

CHLORO PENTAMMINE COBALTIC CHLORIDE, 137. 
Chromates, alkaline earth, 75. 
Chromic chloride modifications, table, 146. 
CHROMIUM, 93. 
CHROMOUS ACETATE, 145. 
Coagulation of colloids, 89. 
Cobalt aluminate, 83. 
Cobalt-ammines, 137. 
Cobalt zincate, 79. 

COLLOIDAL SOLUTIONS, 89. 

Complexion, 115. 
Concentration law, 33. 
Conductivity apparatus, 34. 

CONDUCTIVITY, ELECTRICAL, OF SOLUTIONS, 33. 

Coordination number, 137. 
topper -acetylide, 68. 
'opper refining, 151. 

!ORROSIVE SUBLIMATE, 129. 

irystallization, 113, 116. 

CUPROUS CHLORIDE FROM COPPER, 125. 
! UPROUS OXIDE, PROPORTION BY WEIGHT IN 
WHICH OXYGEN AND COPPER COMBINE TO FORM, 

7. 

CUPROUS SULPHIDE, PROPORTION BY WEIGHT IN 
WHICH SULPHUR AND COPPER COMBINE TO 
FORM, 9. 

Recantation, 109. 
Degree of dissociation, 35, 37. 
Degree of dissociation of silver hydroxide, 95. 
Dichromates, 93. 

Differential method for vapor pressure determi- 
nations, 24. 

Displacement series, Volta, 54, 81, 87, 95, 125. 
)6bereiner's triads, 75. 
Association constant, 40. 



163 



Dissociations of solutions, percentage, table, 36. 
DISSOLVED SOLIDS, EFFECT OF, ON THE BOILING 

POINT, 27. 
DISSOLVED SOLIDS, EFFECT OF, ON THE VAPOR 

PRESSURE, 23. 
Double salts, 111. 
Dropping funnel, use of, 118. 

" E," defined, 5. 

EFFECT OF DISSOLVED SOLIDS ON THE BOILING 

POINT, 27. 
EFFECT OF DISSOLVED SOLIDS ON THE VAPOR 

PRESSURE, 23. 

EFFECT OF TEMPERATURE ON SOLUBILITY, 21. 
ELECTRICAL CONDUCTIVITY OF SOLUTIONS, 33. 
Electrolytic dissociation, 33. 
Electromotive series of the metals, 54. 
Emergency kit, 5. 
Equilibria equations, 37. 
EQUILIBRIA, IONIC, 37. 
Equilibrium, 21. 
EQUILIBRIUM, CHEMICAL, 31. 
Equilibrium constant, 33, 122. 
Equilibrium constant, for water, 40. 
Esterification, 70. 
Esters, 70. 
Etching glass, 51. 
Ethyl alcohol, test, 71. 
Evaporation, 109. 

Filter paper, hardened, 105. 
Fire-polishing, glass, 7. 
Fluorite, 51. 

GASES IN LIQUIDS, SOLUTIONS OF, 19. 
GENERAL INSTRUCTIONS FOR LABORATORY WORK, 

PART i, 3. 
GENERAL INSTRUCTIONS FOR LABORATORY WORK, 

PART II, 97. 

Glass, solubility of, 74. 

Halogen group, 49. 

HALOGEN IONS, PRECIPITATION REACTIONS OF, 
53. 

HALOGENS, AFFINITIES OF, FOR NEGATIVE ELEC- 
TRICITY, 53. 

HALOGENS, OXYGEN ACIDS OF, 55. 

HALOGENS, RELATIVE OXIDIZING POWERS OF THE 
OXYGEN ACIDS OF, 57. 

Hardened filter paper, 105. 

Hexachlor plumbic acid, 88. 

Hirsch funnel, 101. 

HYDRAZINE SULPHATE, 131. 

Hydrazonium hydroxide, 134. 

Hydriodic acid, 51. 



Hydrobromic acid, 51. 

HYDROCARBONS, SATURATED AND UNSATURATED, 

67. 

Hydrochloric acid, 51. 
Hydrofluoric acid, 51. 
HYDROGEN FLUORIDE, BROMIDE, AND IODIDE, 

PREPARATIONS AND PROPERTIES OF, 51. 
HYDROGEN PEROXIDE, 15. 
HYDROGEN SULPHIDE, 41. 
HYDROLYSIS, 43. 

Hydrolysis of orthophosphates, 64. 
Hypobromous acid, 56. 
Hypochlorous acid, 55. 
Hypoiodous acid, 56. 

Indicators, 40. 

Induced reactions, 48. 

Inoculation, 122. 

INSTRUCTIONS FOR LABORATORY WORK, PART i, 

3. 
INSTRUCTIONS FOR LABORATORY WORK, PART n, 

97. 

lodic acid, 56. 

IODIDES, ABSORPTION OF, BY THE STOMACH, 55. 
Iodine clock, 142. 
IODINE, PROPERTIES OF, 49. 
IONIC EQUILIBRIA, 37. 
Ion-product, 40. 
Ions, hydrated, 115. 
Isomorphism, 111. 

Law, concentration, 33. 

LEAD, 87. 

Logarithms, table, 160, 161. 

MAGNESIUM, 77. 
MANGANESE, 91. 
Manometer, 25. 
Marsh's Test, 85. 
MERCURIC CHLORIDE, 129. 
MERCURIC THIOCYANATE, 135. 
MERCURY, 79. 
Metaphosphoric acid, 63. 
Microcosmic salt, 64. 
Mineral chameleon, 91. 
Molar solubility, 22. 
Monochloramine, 131. 

Negative catalysis, 47. 
Nitric acid, concentrated, 113. 

NlTRIC ACID AS AN OXIDIZING AGENT, 61. 

Notes, laboratory, 4, 97. 

-o (termination), 115. 
ORGANIC ACIDS, 71. 



164 



INDEX 



Orthophosphoric acid, 64. 

" Outside Questions," defined, 5. 

OVERVOLTAGE, 81. 

Oxalates, alkaline earth, 75. 

Oxidation, 17. 

OXIDATION AND REDUCTION, 13. 

Oxidizing agents, 13. 

OXIDIZING AGENT, NITRIC ACID AS, 61. 

OXIDIZING POWERS, RELATIVE, OF THE OXYGEN 

ACIDS OF THE HALOGENS, 57. 
OXYGEN ACIDS OF THE HALOGENS, 55. 

Periodic system, 75. 

Periodic table of the elements, 159. 

Permanganates, 91. 

PEROXIDES, 15. 

Pharaoh's serpent eggs, 136. 

Phenolphthalein, 41. 

PHOSPHINE, 63. 

Phosphites, 65. 

PHOSPHORIC ACIDS, 63. 

PHOSPHOROUS ACID, 65. 

Phosphorus, 85. 

Plaited filter, 103. 

Plumbate, ammonium hexachlor, 88. 

Plumbic salts, 88. 

Plumbous salts, 88. 

Polythionic acids, 143. 

Positive catalyst, 48. 

Potassium bromate and iodate, purification of. 

56. 

POTASSIUM IODATE, 141. 
POTASSIUM NITRATE FROM SODIUM NITRATE AND 

POTASSIUM CHLORIDE, 99. 
POTASSIUM THIOCARBONATE, 149. 
PRECIPITATION REACTIONS OF HALOGEN IONS, 53. 
Pressure gauge, 25. 
PROPORTION BY WEIGHT IN WHICH OXYGEN AND 

COPPER COMBINE TO FORM CUPROUS OXIDE, 7. 
PROPORTION BY WEIGHT IN WHICH OXYGEN AND 

SULPHUR COMBINE TO FORM SULPHUR DIOXIDE, 

11. 

PROPORTION BY WEIGHT IN WHICH SULPHUR AND 
COPPER COMBINE TO FORM CUPROUS SULPHIDE, 



' 



Pyrophosphoric acid, 64. 
" Q," defined, 97. 

" R," defined, 3. 
Raschig's method, 131. 
Reagents, use of, 4. 
Red lead, 88. 
Reducing agents, 13. 
REDUCTION, 13. 



Reports, 97. 
Rock salt, 103. 

Saponification, 70. 

SATURATED AND UNSATURATED HYDROCARBONS 

67. 
SELENIUM AND TELLURIUM FROM TELLURIUM 

SLAG, 151. 
Siderite, 111. 
Silicates, 73. 
Silicic acid, 73. 
SILICON, 73. 
SILVER, 95. 
Soda-lime, 7. 

SODIUM CHLORIDE FROM ROCK SALT, 103. 

Sodium hypochlorite, 132. 
Sodium nitrate, 99. 
Solubility curve, 100, 111. 
SOLUBILITY, EFFECT OF TEMPERATURE ON, 21. 
SOLUBILITY OF AIR IN WATER, 19. 
SOLUBILITY OF A SOLID IN A LIQUID, 21. 
Solubility of glass, 74. 
SOLUTIONS, 19, 21, 23, 27. 
SOLUTIONS, COLLOIDAL, 89. 
SOLUTIONS, ELECTRICAL CONDUCTIVITY OF, 33. 
SOLUTIONS, GASES IN LIQUIDS, 19. 
SOLUTIONS, SOLIDS IN LIQUIDS, 21. 
SOLUTIONS, SUPERSATURATED, 23. 
Solutions, use of reagent, 4. 
Standard solution, 41. 
Stannic salts, 87. 
Stannous salts, 87. 
Starch-iodide solution, 15. 
Stibine, 85. 
STRONTIUM, 75. 

STRONTIUM CHLORIDE FROM CELESTITE, 107. 
Sublimation, 129. 
Suction filtration, 101. 
Sulphates, alkaline earth, 75. 
SULPHUR DIOXIDE, PROPORTION BY WEIGHT IN 
WHICH OXYGEN AND SULPHUR COMBINE TO 

FORM, 11. 

SULPHURIC ACID, 45. 
SULPHUROUS ACID, 45. 
Supercooling, 23. 
Superheating, 23. 
Superphosphate, 64. 
SUPERSATURATED SOLUTIONS, 23. 
Supertension, 81. 
Sympathetic reactions, 48. 



TELLURIUM FROM TELLURIUM SLAG, 151. 
TEMPERATURE, EFFECT OF, ON SOLUBILITY, 21. 
Thenard's blue, 83. 



165 



INDEX 



TlN AND LEAD, 87. 

TlTRATTON OF ACIDS AND BASES, 39. 

Triads, Dobereiner's, 75. 

UNSATURATED HYDROCARBONS. SATURATED AND 
67. 

" V," 29. 

VAPOR PRESSURE, EFFECT OF DISSOLVED SOLIDS 

ON, 23. 
Vapor pressure of water, table, 20. 



Velocity constant, 29. 

VELOCITY OF CHEMICAL REACTION, 29. 

Volta Displacement Series, 54. 

Washing by decantation, 109. 

Washing operations, 127. 

Water glass, 74. 

Werner's coordination number, 137. 

Zincate, 79. 

ZlNC, CADMIUM, AND MERCURY, 79 



166 









1932 



15 1932 MAR 20 1936' 



OEC 61933 

SEP 16 1 

MOV 4 1934 
FEB 161935 




^^^ 



YE 02224 




358843 



UNIVERSITY OF CALIFORNIA LIBRARY 



